Inverse Trigonometric Functions
| Formula | Link | Derivative | Domain |
|---|---|---|---|
| [f(x) + g(x)]^\prime | here | f^\prime(x) + g^\prime(x) | |
| [f(x)g(x)]^\prime | here | f^\prime(x)g(x) + f(x)g^\prime(x) | |
| \left[\dfrac{f(x)}{g(x)} \right]^\prime | here | \dfrac{f^\prime(x)g(x) - f(x)g^\prime(x)}{g(x)^2} | |
| [f(g(x))]^\prime | here | f^\prime(g(x)) g^\prime(x) |
| Formula | Link | Derivative | Domain |
|---|---|---|---|
| x^n | here | n \, x^{n-1} | varies |
| e^x | here | e^x | -\infty < x < \infty |
| a^x = e^{\ln(a)x} | here | \ln(a) \, a^x | -\infty < x < \infty, a > 0 |
| \ln(x) | here | \dfrac{1}{x} | 0 < x < \infty |
| \log_a(x) = \dfrac{\ln(x)}{\ln(a)} | here | \dfrac{1}{x \, \ln(a)} | 0 < x < \infty, a > 0 |
| Formula | Link | Derivative | Domain |
|---|---|---|---|
| \sin(x) | here | \cos(x) | -\infty < x < \infty |
| \cos(x) | here | -\sin(x) | -\infty < x < \infty |
| \tan(x) | here | \frac{1}{\cos(x)^2} = \sec(x)^2 | x \ne \pi + \pi \,n, n \in \mathbb{Z} |
| \cot(x) | here | -\frac{1}{\sin(x)^2} = -\csc(x)^2 | x \ne \frac{\pi}{2} + \pi \,n, n \in \mathbb{Z} |
| \csc(x) | here | -\frac{\cos(x)}{\sin(x)}\frac{1}{\sin(x)} = -\cot(x)\csc(x) | x \ne \pi + \pi \,n, n \in \mathbb{Z} |
| \sec(x) | here | \frac{\tan(x)}{\cos(x)} = \sec(x)\tan(x) | x \ne \frac{\pi}{2} + \pi \,n, n \in \mathbb{Z} |
| Formula | Link | Derivative | Domain |
|---|---|---|---|
| \arcsin(x) | here | \frac{1}{\sqrt{1 - x^2}} | -1 < x < 1 |
| \arccos(x) | here | -\frac{1}{\sqrt{1 - x^2}} | -1 < x < 1 |
| \arctan(x) | here | \frac{1}{1 + x^2} | -\infty < x < \infty |
| \operatorname{arccot}(x) | here | -\frac{1}{1+x^2} | -\infty < x < \infty |
| \operatorname{arccsc}(x) | here | -\frac{1}{|x| \sqrt{x^2 -1}} | x \in (-\infty, 1)\, \cup \, (1,\infty) |
| \operatorname{arcsec}(x) | here | \frac{1}{|x|\sqrt{x^2 -1}} | x \in (-\infty, 1)\, \cup \, (1,\infty) |
| Formula | Link | Derivative | Domain |
|---|---|---|---|
| \sinh(x) | here | \cosh(x) | -\infty < x < \infty |
| \cosh(x) | here | \sinh(x) | -\infty < x < \infty |
| \tanh(x) | here | \text{sech}^2(x) | -\infty < x < \infty |
| \coth(x) | here | -\text{csch}^2(x) | x \ne 0 |
| \text{csch}(x) | here | -\coth(x)\text{csch}(x) | x \ne 0 |
| \text{sech}(x) | here | -\tanh(x)\text{sech}(x) | -\infty < x < \infty |
| Formula | Link | Derivative | Domain |
|---|---|---|---|
| \text{arcsinh}(x) | here | \frac{1}{\sqrt{x^2+1}} | -\infty < x < \infty |
| \text{arccosh}(x) | here | \frac{1}{\sqrt{x^2-1}} | x > 1 |
| \text{arctanh}(x) | here | \frac{1}{1-x^2} | |x| < 1 |
| \text{arccoth}(x) | here | \frac{1}{1-x^2} | |x| > 1 |
| \text{arccsch}(x) | here | -\frac{1}{|x|\sqrt{x^2+1}} | x \ne 0 |
| \text{arcsech}(x) | here | -\frac{1}{x\sqrt{1-x^2}} | 0 < x < 1 |
Let h(x) = f(x) + g(x). Applying the definition of the derivative:
\begin{aligned} h^\prime(x) &= \lim_{\Delta x \to 0} \frac{h(x+\Delta x) - h(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{[f(x+\Delta x) + g(x+\Delta x)] - [f(x) + g(x)]}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x) + g(x+\Delta x) - g(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} + \lim_{\Delta x \to 0} \frac{g(x+\Delta x) - g(x)}{\Delta x} \\ &= f^\prime(x) + g^\prime(x) \end{aligned}
Let h(x) = f(x)g(x). Applying the definition of the derivative and adding and subtracting f(x+\Delta x)g(x) in the numerator:
\begin{aligned} h^\prime(x) &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)g(x+\Delta x) - f(x)g(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)g(x+\Delta x) - f(x+\Delta x)g(x) + f(x+\Delta x)g(x) - f(x)g(x)}{\Delta x} \\ &= \lim_{\Delta x \to 0} \left[ f(x+\Delta x) \frac{g(x+\Delta x) - g(x)}{\Delta x} + g(x) \frac{f(x+\Delta x) - f(x)}{\Delta x} \right] \\ &= \left(\lim_{\Delta x \to 0} f(x+\Delta x)\right) \left(\lim_{\Delta x \to 0} \frac{g(x+\Delta x) - g(x)}{\Delta x}\right) + \left(\lim_{\Delta x \to 0} g(x)\right) \left(\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}\right) \\ &= f(x)g^\prime(x) + g(x)f^\prime(x) \end{aligned}
Let h(x) = \frac{f(x)}{g(x)}. Applying the definition of the derivative:
\begin{aligned} h^\prime(x) &= \lim_{\Delta x \to 0} \frac{\frac{f(x+\Delta x)}{g(x+\Delta x)} - \frac{f(x)}{g(x)}}{\Delta x} \\ &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)g(x) - f(x)g(x+\Delta x)}{\Delta x \cdot g(x+\Delta x)g(x)} \\ \end{aligned}
Now, we add and subtract f(x)g(x) in the numerator:
\begin{aligned} h^\prime(x) &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)g(x) - f(x)g(x) - f(x)g(x+\Delta x) + f(x)g(x)}{\Delta x \cdot g(x+\Delta x)g(x)} \\ &= \lim_{\Delta x \to 0} \frac{g(x)\frac{f(x+\Delta x)-f(x)}{\Delta x} - f(x)\frac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x+\Delta x)g(x)} \\ &= \frac{\lim_{\Delta x \to 0} g(x)\frac{f(x+\Delta x)-f(x)}{\Delta x} - \lim_{\Delta x \to 0} f(x)\frac{g(x+\Delta x)-g(x)}{\Delta x}}{\lim_{\Delta x \to 0} g(x+\Delta x)g(x)} \\ &= \frac{g(x)f^\prime(x) - f(x)g^\prime(x)}{[g(x)]^2} \end{aligned}
Let h(x) = f(g(x)). By the definition of the derivative:
h^\prime(x) = \lim_{\Delta x \to 0} \frac{f(g(x + \Delta x)) - f(g(x))}{\Delta x}
Let \Delta g = g(x + \Delta x) - g(x). As \Delta x \to 0, \Delta g \to 0. We multiply the numerator and denominator by \Delta g:
\begin{aligned} h^\prime(x) &= \lim_{\Delta x \to 0} \frac{f(g(x) + \Delta g) - f(g(x))}{\Delta g} \cdot \frac{\Delta g}{\Delta x} \\ &= \lim_{\Delta g \to 0} \frac{f(g(x) + \Delta g) - f(g(x))}{\Delta g} \cdot \lim_{\Delta x \to 0} \frac{g(x + \Delta x) - g(x)}{\Delta x} \\ &= f^\prime(g(x)) \cdot g^\prime(x) \end{aligned}
Starting with a polynomial of second degree (e.g. f(x) = x^2):
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{(x + h)^2 - x^2}{h} \\ & = \lim\limits_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \\ & = \lim\limits_{h \to 0} \frac{ h (2x + h) }{h} \\ & = \lim\limits_{h \to 0} (2x + h) \\ & = 2x \end{aligned}
In general, for a polynomial f(x) = x^n, the derivative is f^\prime(x) = n\, x^{n-1}, known as power rule and the demonstration is similar.
From the definition of e^x as (taking n=\frac{1}{h} and x=1):
e^x = \lim\limits_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = \lim\limits_{h \to 0} (1 + xh)^{\frac{1}{h}}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{e^{x + h} - e^x}{h} \\ & = \lim\limits_{h \to 0} \frac{e^xe^h - e^x}{h} \\ & = \lim\limits_{h \to 0} \frac{ e^x (e^h - 1) }{h} \\ & = e^x \lim\limits_{h \to 0} \frac{e^h -1}{h} \\ & = e^x \lim\limits_{h \to 0} \frac{\left((1 + h)^{\frac{1}{h}}\right)^h - 1}{h} \\ & = e^x \lim\limits_{h \to 0} \frac{\left( 1 + h \right) - 1}{h} \\ & = e^x \end{aligned}
From the definition of e^x as (taking n=\frac{1}{h} and x=1):
e^x = \lim\limits_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = \lim\limits_{h \to 0} (1 + xh)^{\frac{1}{h}}
From the definition of \ln(a) as (considering a^h - 1 =m and therefore h = \frac{\ln(m+1)}{\ln(a)}):
\lim\limits_{h \rightarrow 0} \frac{a^h-1}{h} = \lim\limits_{h \rightarrow 0} m \frac{\ln(a)}{\ln(m+1)} =\lim\limits_{h \rightarrow 0} \frac{\ln(a)}{\ln(1+m)^{1/m}} = \frac{\ln(a)}{\ln(e)} = \ln{a}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{a^{x + h} - a^x}{h} \\ & = a^x \lim\limits_{h \rightarrow 0} \frac{a^h-1}{h} \\ & = a^x \ln(a) \end{aligned}
From the definition of e^x as (taking n=\frac{1}{h}):
e^{\frac{1}{x}} = \lim\limits_{n \to \infty} \left( 1 + \frac{1}{x} \frac{1}{n} \right)^n = \lim\limits_{h \to 0} (1 + \frac{h}{x})^{\frac{1}{h}}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\ln(x + h) - \ln(x)}{h} \\ & = \lim\limits_{h \to 0} \frac{\ln \left( \frac{x+h}{x} \right)}{h} \\ & = \lim\limits_{h \to 0} \ln \left( \frac{x+h}{x} \right)^{\frac{1}{h}} \\ & = \lim\limits_{h \to 0} \ln \left( 1 + \frac{h}{x} \right)^{\frac{1}{h}} \\ & = \ln \left( e^{\frac{1}{x}} \right) \\ & = \frac{1}{x} \end{aligned}
From the property of the logarithm:
log_a(x) = \frac{ln(x)}{ln(a)}
and from the proof of the derivatives of \ln(x).
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\log_a(x + h) - \log_a(x)}{h} \\ & = \lim\limits_{h \to 0} \frac{\frac{\ln(x + h)}{\ln(a)} - \frac{\ln(x)}{\ln(a)}}{h} \\ & = \frac{1}{\ln(a)} \lim\limits_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h} \\ & = \frac{1}{\ln(a)} \frac{1}{x} \end{aligned}
Using the identity \sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B):
and:
\begin{gathered} \lim\limits_{h \to 0} \frac{(1 - \cos(h))}{h} = 0 \\ \lim\limits_{h \to 0} \frac{\sin(h)}{h} = 1 \end{gathered}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\sin(x + h) - \sin(x)}{h} \\ & = \lim\limits_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} \\ & = \lim\limits_{h \to 0} \frac{\sin(x)(1 - \cos(h)) + \cos(x)\sin(h)}{h} \\ & = \lim\limits_{h \to 0} \frac{\sin(x)(1 - \cos(h))}{h} + \lim\limits_{h \to 0} \frac{\cos(x)\sin(h)}{h} \\ & = \cos(x) \\ \end{aligned}
Using the identity \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B):
and
\begin{gathered} \lim\limits_{h \to 0} \frac{(1 - \cos(h))}{h} = 0 \\ \lim\limits_{h \to 0} \frac{\sin(h)}{h} = 1 \end{gathered}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\cos(x + h) - \cos(x)}{h} \\ & = \lim\limits_{h \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h} \\ & = \lim\limits_{h \to 0} \frac{\cos(x)(\cos(h) - 1) - \sin(x)\sin(h)}{h} \\ & = \lim\limits_{h \to 0} \frac{-\cos(x)(1 - \cos(h)) - \sin(x)\sin(h)}{h} \\ & = -\sin(x) \\ \end{aligned}
Using the definition of \tan(x) = \frac{sin(x)}{\cos(x)} and the identity \sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)
and
\begin{gathered} \lim\limits_{h \to 0} \frac{(1 - \cos(h))}{h} = 0 \\ \lim\limits_{h \to 0} \frac{\sin(h)}{h} = 1 \end{gathered}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\tan(x + h) - \tan(x)}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin(x)}{\cos(x)}}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{\sin(x+h)\cos(x) - \cos(x+h) \cos(x)}{\cos(x+h) \cos(x)}}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\sin(x+h-x)}{h \cos(x+h) \cos(x)} \\ & = \lim\limits_{h \rightarrow 0} \frac{\sin(h)}{h \cos(x+h) \cos(x)} \\ & = \lim\limits_{h \rightarrow 0} \frac{\sin(h)}{h} \lim\limits_{h \rightarrow 0} \frac{1}{\cos(x+h) \cos(x)} \\ & = \lim\limits_{h \rightarrow 0} \frac{1}{\cos(x+h) \cos(x)} \\ & = \frac{1}{\cos(x) \cos(x)} = \frac{1}{\cos(x)^2} = \sec(x)^2 \end{aligned}
Using the identity \sin(A - B) = \sin(A)\cos(B) - \sin(B)\cos(A):
and
\begin{gathered} \lim\limits_{h \to 0} \frac{\sin(h)}{h} = 1 \end{gathered}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\cot(x + h) - \cot(x)}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{\cos(x+h)}{\sin(x+h)} - \frac{\cos(x)}{\sin(x)}}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{\cos(x+h)\sin(x) - \sin(x+h) \cos(x)}{\sin(x+h) \sin(x)}}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\sin(x-x-h)}{h \sin(x+h) \sin(x)} \\ & = \lim\limits_{h \rightarrow 0} \frac{\sin(-h)}{h \sin(x+h) \sin(x)} \\ & = - \lim\limits_{h \rightarrow 0} \frac{\sin(h)}{h} \lim\limits_{h \rightarrow 0} \frac{1}{\cos(x+h) \cos(x)} \\ & = - \lim\limits_{h \rightarrow 0} \frac{1}{\sin(x+h) \sin(x)} \\ & = - \frac{1}{\sin(x) \sin(x)} = \frac{1}{\sin(x)^2} = - \csc(x)^2 \end{aligned}
Using the definition of derivative of \frac{\mathrm{d}\sin(x)}{\mathrm{d}x} = \cos(x) and applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\csc(x + h) - \csc(x)}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{1}{\sin(x+h)} - \frac{1}{\sin(x)}}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{\sin(x) - \sin(x+h)}{\sin(x+h) \sin(x)}}{h} \\ & = - \lim\limits_{h \rightarrow 0} \frac{\sin(x+h) - \sin(x)}{h} \lim\limits_{h \rightarrow 0}\frac{1}{\sin(x+h) \sin(x)} \\ & = - \cos(x) \lim\limits_{h \rightarrow 0}\frac{1}{\sin(x+h) \sin(x)} \\ & = - \cos(x) \frac{1}{\sin(x) \sin(x)} = - \frac{\cos(x)}{\sin(x)} \frac{1}{\sin(x)} \\ & = - \cot(x) \csc(x) \\ \end{aligned}
Using the definition of derivative of \frac{\mathrm{d}\cos(x)}{\mathrm{d}x} = -\sin(x) and applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\sec(x + h) - \sec(x)}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{1}{\cos(x+h)} - \frac{1}{\cos(x)}}{h} \\ & = \lim\limits_{h \rightarrow 0} \frac{\frac{\cos(x) - \cos(x+h)}{\cos(x+h) \cos(x)}}{h} \\ & = - \lim\limits_{h \rightarrow 0} \frac{\cos(x+h) - \cos(x)}{h} \lim\limits_{h \rightarrow 0}\frac{1}{\cos(x+h) \cos(x)} \\ & = \sin(x) \lim\limits_{h \rightarrow 0}\frac{1}{\cos(x+h) \cos(x)} \\ & = \sin(x) \frac{1}{\cos(x) \cos(x)} = \frac{\sin(x)}{\cos(x)} \frac{1}{\cos(x)} \\ & = \tan(x) \sec(x) \\ \end{aligned}
Using the identity \sin(A) - \sin(B) = 2 \sin(\frac{A-B}{2})\cos(\frac{A+B}{2}) and performing the substitution:
\begin{aligned} & u = \arcsin(x+h) \\ & v = \arcsin(x+h) \\ & \sin(u) = \sin(\arcsin(x+h)) = x + h \\ & \sin(v) = \sin(\arcsin(x)) = x \\ & \Rightarrow h = \sin(u) - \sin(v) \\ \end{aligned}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\arcsin(x + h) - \arcsin(x)}{h} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\sin(u) - \sin(v)} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{2 \sin(\frac{u-v}{2})\cos(\frac{u+v}{2})} \\ & = \lim\limits_{u \rightarrow v} \frac{\frac{u - v}{2}}{\sin(\frac{u-v}{2})}\lim\limits_{u \rightarrow v} \frac{1}{\cos(\frac{u+v}{2})} \\ & \lim\limits_{u \rightarrow v} \frac{1}{\cos(\frac{u+v}{2})} \\ & \frac{1}{\cos(v)} = \frac{1}{\sqrt{1-\sin(v)^2}} = \frac{1}{\sqrt{1-\sin(\arcsin(x))^2}} \\ & = \frac{1}{\sqrt{1 - x^2}} \end{aligned}
Using the identity \cos(A) - \cos(B) = -2 \sin(\frac{A-B}{2})\sin(\frac{A+B}{2}) and performing the substitution:
\begin{aligned} & u = \arccos(x+h) \\ & v = \arccos(x+h) \\ & \cos(u) = \cos(\arccos(x+h)) = x + h \\ & \cos(v) = \cos(\arccos(x)) = x \\ & \Rightarrow h = \cos(u) - \cos(v) \\ \end{aligned}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\arccos(x + h) - \arccos(x)}{h} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\cos(u) - \cos(v)} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{-2 \sin(\frac{u-v}{2})\sin(\frac{u+v}{2})} \\ & = - \lim\limits_{u \rightarrow v} \frac{\frac{u - v}{2}}{\sin(\frac{u-v}{2})}\lim\limits_{u \rightarrow v} \frac{1}{\sin(\frac{u+v}{2})} \\ & -\lim\limits_{u \rightarrow v} \frac{1}{\sin(\frac{u+v}{2})} \\ & -\frac{1}{\sin(v)} = -\frac{1}{\sqrt{1-\cos(v)^2}} = - \frac{1}{\sqrt{1-\cos(\arccos(x))^2}} \\ & = - \frac{1}{\sqrt{1 - x^2}} \end{aligned}
Using the identity \sin(A - B) = \sin(A)\cos(B) - \sin(B)\cos(A), the formula
\begin{aligned} & y = \arctan(x) \Rightarrow x = \tan(y) = \frac{\sin(y)}{\cos(y)} \\ & x^2 = \frac{\sin(y)^2}{\cos(y)^2} \\ & 1 + x^2 = \frac{\sin(y)^2}{\cos(y)^2} + 1 = \frac{\sin(y)^2 + \cos(y)^2}{\cos(y)^2} = \frac{1}{\cos(y)^2}\\ & 1 + x^2 = \frac{1}{\cos(\arctan(x))^2} \end{aligned}
and performing the substitution:
\begin{aligned} & u = \arctan(x+h) \\ & v = \arctan(x+h) \\ & \cot(u) = \tan(\arctan(x+h)) = x + h \\ & \cos(v) = \tan(\arctan(x)) = x \\ & \Rightarrow h = \tan(u) - \tan(v) \\ \end{aligned}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\arctan(x + h) - \arctan(x)}{h} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\tan(u) - \tan(v)} = \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\sin(u)}{\cos(u)} - \frac{\sin(v)}{\cos(v)}} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\sin(u)\cos(v) - \sin(v)\cos(u)}{\cos(u)\cos(v)}} = \lim\limits_{u \rightarrow v} \frac{u - v}{\sin(u-v)} \lim\limits_{u \rightarrow v} \cos(u)\cos(v) = \lim\limits_{u \rightarrow v} \cos(u)\cos(v) \\ & = \cos(v)^2 = \cos(\arctan(x))^2 = \frac{1}{1+x^2} \end{aligned}
Using the identity \cos(A) - \cos(B) = -2 \sin(\frac{A-B}{2})\sin(\frac{A+B}{2}) and performing the substitution:
Using the identity \sin(A - B) = \sin(A)\cos(B) - \sin(B)\cos(A), the formula
\begin{aligned} & y = \operatorname{arccot}(x) \Rightarrow x = \cot(y) = \frac{\cos(y)}{\sin(y)} \\ & x^2 = \frac{\cos(y)^2}{\sin(y)^2} \\ & 1 + x^2 = \frac{\cos(y)^2}{\sin(y)^2} + 1 = \frac{\cos(y)^2 + \sin(y)^2}{\sin(y)^2} = \frac{1}{\sin(y)^2}\\ & 1 + x^2 = \frac{1}{\sin(\operatorname{arccot}(x))^2} \end{aligned}
and performing the substitution:
\begin{aligned} & u = \operatorname{arccot}(x+h) \\ & v = \operatorname{arccot}(x+h) \\ & \cot(u) = \cot(\operatorname{arccot}(x+h)) = x + h \\ & \cos(v) = \cot(\operatorname{arccot}(x)) = x \\ & \Rightarrow h = \cot(u) - \cot(v) \\ \end{aligned}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\operatorname{arccot}(x + h) - \operatorname{arccot}(x)}{h} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\cot(u) - \cot(v)} = \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\cos(u)}{\sin(u)} - \frac{\cos(v)}{\sin(v)}} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\sin(v)\cos(u) - \sin(u)\cos(v)}{\sin(u)\sin(v)}} = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\sin(u)\cos(v) -\sin(v)\cos(u)}{\sin(u)\sin(v)}} \\ & - \lim\limits_{u \rightarrow v} \frac{u - v}{\sin(u-v)}\lim\limits_{u \rightarrow v} \sin(u)\sin(v) = \lim\limits_{u \rightarrow v} \sin(u)\sin(v) \\ & = -\sin(v)^2 = -\sin(\operatorname{arccot})^2 = -\frac{1}{1+x^2} \end{aligned}
Using the identity \sin(A) - \sin(B) = 2 \sin(\frac{A-B}{2})\cos(\frac{A+B}{2}), the formula
\cot(x)^2 = \frac{\cos(x)^2}{\sin(x)^2} = \frac{1 - \sin(x)^2}{\sin(x)} = \csc(x)^2 - 1
and performing the substitution:
\begin{aligned} & u = \operatorname{arccsc}(x+h) \\ & v = \operatorname{arccsc}(x+h) \\ & \sec(u) = \sec(\operatorname{arccsc}(x+h)) = x + h \\ & \sec(v) = \sec(\operatorname{arccsc}(x)) = x \\ & \Rightarrow h = \csc(u) - \csc(v) \end{aligned}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\operatorname{arcsc}(x + h) - \operatorname{arcsc}(x)}{h} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\csc(u) - \csc(v)} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{1}{\sin(u)} - \frac{1}{\sin(v)}} = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{1}{\sin(v)} - \frac{1}{\sin(u)}} = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\sin(u) - \sin(v)}{\sin(u)\sin(v)}} \\ & = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{2 \sin(\frac{u-v}{2})\cos(\frac{u+v}{2})}{\sin(u)\sin(v)}} \\ & = - \lim\limits_{u \rightarrow v} \frac{\frac{u - v}{2}}{\sin(\frac{u-v}{2})} \lim\limits_{u \rightarrow v} \frac{1}{\frac{\cos(\frac{u+v}{2})}{\sin(u)\sin(v)}} = - \lim\limits_{u \rightarrow v} \frac{1}{\frac{\cos(\frac{u+v}{2})}{\sin(u)\sin(v)}} \\ & = - \frac{1}{\csc(v) \cot(v)} \end{aligned}
Taking the square and imposing that it is always negative:
\begin{aligned} & \left(\frac{\mathrm{d}f(x)}{\mathrm{d}x} \right)^2 = \frac{1}{\csc(v)^2 \cot(v)^2} = \frac{1}{\csc(v)^2 (\csc(v)^2 - 1)} = \frac{1}{x^2 (x^2 - 1)} \\ & \frac{\mathrm{d}f(x)}{\mathrm{d}x} = - \frac{1}{|x| \sqrt{x^2 - 1}} \end{aligned}
Using the identity \cos(A) - \cos(B) = -2 \sin(\frac{A-B}{2})\sin(\frac{A+B}{2}), the formula
\tan(x)^2 = \frac{\sin(x)^2}{\cos(x)^2} = \frac{1 - \cos(x)^2}{\cos(x)} = \sec(x)^2 - 1
and performing the substitution:
\begin{aligned} & u = \operatorname{arcsec}(x+h) \\ & v = \operatorname{arcsec}(x+h) \\ & \sec(u) = \sec(\operatorname{arcsec}(x+h)) = x + h \\ & \sec(v) = \sec(\operatorname{arcsec}(x)) = x \\ & \Rightarrow h = \sec(u) - \sec(v) \end{aligned}
Applying the definition:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \lim\limits_{h \rightarrow 0} \frac{\operatorname{arcsec}(x + h) - \operatorname{arcsec}(x)}{h} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\sec(u) - \sec(v)} \\ & = \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{1}{\cos(u)} - \frac{1}{\cos(v)}} = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{1}{\cos(v)} - \frac{1}{\cos(u)}} = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{\cos(u) - \cos(v)}{\cos(u)\cos(v)}} \\ & = - \lim\limits_{u \rightarrow v} \frac{u - v}{\frac{-2 \sin(\frac{u-v}{2})\sin(\frac{u+v}{2})}{\cos(u)\cos(v)}} \\ & = \lim\limits_{u \rightarrow v} \frac{\frac{u - v}{2}}{\sin(\frac{u-v}{2})} \lim\limits_{u \rightarrow v} \frac{1}{\frac{\sin(\frac{u+v}{2})}{\cos(u)\cos(v)}} = \lim\limits_{u \rightarrow v} \frac{1}{\frac{\sin(\frac{u+v}{2})}{\cos(u)\cos(v)}} \\ & = \frac{1}{\sec(v) \tan(v)} \end{aligned}
Taking the square and imposing that it is always positive:
\begin{aligned} & \left(\frac{\mathrm{d}f(x)}{\mathrm{d}x} \right)^2 = \frac{1}{\sec(v)^2 \tan(v)^2} = \frac{1}{\sec(v)^2 (\sec(v)^2 - 1)} = \frac{1}{x^2 (x^2 - 1)} \\ & \frac{\mathrm{d}f(x)}{\mathrm{d}x} = \frac{1}{|x| \sqrt{x^2 - 1}} \end{aligned}
Of course, here is the equivalent information for hyperbolic functions in a markdown format.
Using the definition \sinh(x) = \frac{e^x - e^{-x}}{2} and the derivatives of exponential functions:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{e^x - e^{-x}}{2} \right) \\ & = \frac{1}{2} \left( \frac{\mathrm{d}}{\mathrm{d}x}e^x - \frac{\mathrm{d}}{\mathrm{d}x}e^{-x} \right) \\ & = \frac{1}{2} (e^x - (-e^{-x})) \\ & = \frac{e^x + e^{-x}}{2} \\ & = \cosh(x) \end{aligned}
Using the definition \cosh(x) = \frac{e^x + e^{-x}}{2} and the derivatives of exponential functions:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{e^x + e^{-x}}{2} \right) \\ & = \frac{1}{2} \left( \frac{\mathrm{d}}{\mathrm{d}x}e^x + \frac{\mathrm{d}}{\mathrm{d}x}e^{-x} \right) \\ & = \frac{1}{2} (e^x - e^{-x}) \\ & = \frac{e^x - e^{-x}}{2} \\ & = \sinh(x) \end{aligned}
Using the definition \tanh(x) = \frac{\sinh(x)}{\cosh(x)}, the derivatives of \sinh(x) and \cosh(x), and the quotient rule:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\sinh(x)}{\cosh(x)} \right) \\ & = \frac{\frac{\mathrm{d}\sinh(x)}{\mathrm{d}x}\cosh(x) - \sinh(x)\frac{\mathrm{d}\cosh(x)}{\mathrm{d}x}}{\cosh^2(x)} \\ & = \frac{\cosh(x)\cosh(x) - \sinh(x)\sinh(x)}{\cosh^2(x)} \\ & = \frac{\cosh^2(x) - \sinh^2(x)}{\cosh^2(x)} \end{aligned}
Using the identity \cosh^2(x) - \sinh^2(x) = 1:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{1}{\cosh^2(x)} = \text{sech}^2(x) \end{aligned}
Using the definition \coth(x) = \frac{\cosh(x)}{\sinh(x)}, the derivatives of \sinh(x) and \cosh(x), and the quotient rule:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\cosh(x)}{\sinh(x)} \right) \\ & = \frac{\frac{\mathrm{d}\cosh(x)}{\mathrm{d}x}\sinh(x) - \cosh(x)\frac{\mathrm{d}\sinh(x)}{\mathrm{d}x}}{\sinh^2(x)} \\ & = \frac{\sinh(x)\sinh(x) - \cosh(x)\cosh(x)}{\sinh^2(x)} \\ & = \frac{\sinh^2(x) - \cosh^2(x)}{\sinh^2(x)} \end{aligned}
Using the identity \sinh^2(x) - \cosh^2(x) = -1:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = -\frac{1}{\sinh^2(x)} = -\text{csch}^2(x) \end{aligned}
Using the definition \text{csch}(x) = \frac{1}{\sinh(x)}, the derivative of \sinh(x), and the chain rule:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} (\sinh(x))^{-1} \\ & = -1(\sinh(x))^{-2} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\sinh(x) \\ & = -\frac{1}{\sinh^2(x)} \cdot \cosh(x) \\ & = -\frac{\cosh(x)}{\sinh(x)} \frac{1}{\sinh(x)} \\ & = -\coth(x)\text{csch}(x) \end{aligned}
Using the definition \text{sech}(x) = \frac{1}{\cosh(x)}, the derivative of \cosh(x), and the chain rule:
\begin{aligned} \frac{\mathrm{d}f(x)}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} (\cosh(x))^{-1} \\ & = -1(\cosh(x))^{-2} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\cosh(x) \\ & = -\frac{1}{\cosh^2(x)} \cdot \sinh(x) \\ & = -\frac{\sinh(x)}{\cosh(x)} \frac{1}{\cosh(x)} \\ & = -\tanh(x)\text{sech}(x) \end{aligned}
Let y = \text{arcsinh}(x). Then x = \sinh(y). Differentiating with respect to x:
\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \sinh(y) \\ 1 & = \cosh(y) \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{\cosh(y)} \end{aligned}
Using the identity \cosh^2(y) - \sinh^2(y) = 1, we have \cosh(y) = \sqrt{1 + \sinh^2(y)}. Since \cosh(y) is always positive, we take the positive root. Substituting back x = \sinh(y):
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1 + x^2}}
Let y = \text{arccosh}(x). Then x = \cosh(y). Differentiating with respect to x:
\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \cosh(y) \\ 1 & = \sinh(y) \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{\sinh(y)} \end{aligned}
Using the identity \cosh^2(y) - \sinh^2(y) = 1, we have \sinh(y) = \sqrt{\cosh^2(y) - 1}. For x > 1, the range of \text{arccosh}(x) is y > 0, where \sinh(y) is positive. Substituting back x = \cosh(y):
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{x^2 - 1}}
Let y = \text{arctanh}(x). Then x = \tanh(y). Differentiating with respect to x:
\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \tanh(y) \\ 1 & = \text{sech}^2(y) \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{\text{sech}^2(y)} \end{aligned}
Using the identity \text{sech}^2(y) = 1 - \tanh^2(y), and substituting back x = \tanh(y):
\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 - x^2}
Let y = \text{arccoth}(x). Then x = \coth(y). Differentiating with respect to x:
\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \coth(y) \\ 1 & = -\text{csch}^2(y) \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = -\frac{1}{\text{csch}^2(y)} \end{aligned}
Using the identity \text{coth}^2(y) - 1 = \text{csch}^2(y), and substituting back x = \coth(y):
\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x^2 - 1} = \frac{1}{1 - x^2}
Let y = \text{arccsch}(x). Then x = \text{csch}(y). Differentiating with respect to x:
\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \text{csch}(y) \\ 1 & = -\text{csch}(y)\coth(y) \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = -\frac{1}{\text{csch}(y)\coth(y)} \end{aligned}
Using the identity \coth^2(y) = 1 + \text{csch}^2(y), we have \coth(y) = \pm\sqrt{1 + \text{csch}^2(y)}. Substituting back x = \text{csch}(y):
\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x (\pm\sqrt{1+x^2})} = -\frac{1}{|x|\sqrt{1+x^2}}
Let y = \text{arcsech}(x). Then x = \text{sech}(y). Differentiating with respect to x:
\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}x} & = \frac{\mathrm{d}}{\mathrm{d}x} \text{sech}(y) \\ 1 & = -\text{sech}(y)\tanh(y) \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = -\frac{1}{\text{sech}(y)\tanh(y)} \end{aligned}
Using the identity \tanh^2(y) = 1 - \text{sech}^2(y), we have \tanh(y) = \sqrt{1 - \text{sech}^2(y)} for 0 < x < 1, which corresponds to y > 0. Substituting back x = \text{sech}(y):
\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x\sqrt{1-x^2}}