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Common derivatives

The derivative of the power function defined for $$\alpha\in\R$$ is given by the formula $$$(x^\alpha)' = \alpha x^{\alpha -1}$$$

First proof: For $$\alpha\in\N$$, this can be proved by induction and by the rule for the derivative of the multiplication. The property is true for $$\beta = 0$$, as the derivative of a constant is 0. If we assume the property to be true for $$\beta$$, we have $$$(x^{\beta+1})' = (x x^{\beta})' = x^{\beta} + \beta x x^{\beta-1} = (\beta+1) x^\beta.$$$ So the property is true for $$\beta+1$$. This proves the result by induction.

Second proof: For $$\alpha\in\R$$, we use the derivative of $$e^x$$ and $$\ln(x)$$ that will be recalled later. $$$(x^{\alpha})' = (e^{\alpha\ln x})' = \alpha x^{-1} e^{\alpha\ln x} = \alpha x^{\alpha-1}.$$$

The derivative of the exponential function, defined for $$\alpha>0$$, is given by the formula \begin{gather*} (e^x)' = e^x,\\ (\alpha^x)' = \alpha^x\,\ln\alpha \end{gather*}

Proof: You will have to remember the derivative of $$e^x$$! (The proof depends on how the function is defined and is beyond the syllabus.)

For the general exponential function, we have $$$ (\alpha^x)' = (e^{x\ln\alpha})' = \alpha^x\,\ln\alpha$$$

The derivative of the logarithmic function, defined for $$\alpha \gt 0$$, is given by the formula (for $$x\gt0$$) $$$ (\ln x)' = \frac{1}{x},\qquad (\log_\alpha x)'=\frac{1}{x\ln\alpha} $$$

When $$x\ne 0$$, it extends to $$$ (\ln\vert x\vert )' = \frac{1}{x},\qquad (\log_\alpha\vert x\vert)'=\frac{1}{x\ln\alpha} $$$

Proof: The logarithm function $$\ln$$ is the inverse of the exponential $$e^x$$. We can apply the general formula for the derivative of the inverse. $$$ (\ln x)' = \big((e^x)^{-1}\big)' = \frac{1}{e^{\ln x}} =\frac{1}{x}$$$

When $$x$$ is negative, we use that $$\vert x\vert= -x$$. This gives $$$ (\ln\vert x\vert )' = \big(\ln(- x)\big )' = \frac{-1}{-x} = \frac{1}{x}.$$$

For the general logarithm with base $$\alpha$$, we use $$$ (\log_\alpha x)= \frac{\ln x}{\ln\alpha}.$$$

The derivatives of the sine and the cosine functions are given by the formula $$$ (\sin x)' = \cos x, \qquad (\cos x)' = -\sin x. $$$

Proofs: For $$\sin x$$, you will have to remember the derivative (the proof depends on how the function is defined). If you are not sure of the sign, note that the function is increasing at 0 so the derivative at 0 must be positive.

For $$\cos x$$, we differentiate the identity $$\cos x = \sin(\pi/2-x)$$. We get $$$ (\cos x)' = \sin'(\pi/2-x) = -\cos(\pi/2-x) = -\sin x $$$

The derivatives of the advanced trigonometric functions are given by the formulae \begin{gather*} (\tan x)' = \frac{1}{\cos^2x} = 1+\tan^2 x,\\ (\cot x)' = -\frac{1}{\sin^2x} = -1-\cot^2 x,\\ (\sec x)' = \sec x\tan x, \quad (\csc x)' = -\csc x\cot x \end{gather*}

Proofs: We use the expression for the derivative of the quotient.

\begin{align*} (\tan x)' &=\big(\frac{\sin x}{\cos x}\big)' = \frac{\sin'x\cos x - \cos'x \sin x}{\cos^2 x}\\ &= \frac{\cos^2x + \sin^2x}{\cos^2 x} = \frac{1}{\cos^2 x},\\ (\cot x)' &= \big(\tan(\frac{\pi}{2}-x)\big)' = \frac{-1}{\cos^2(\frac{\pi}{2}-x)} = \frac{-1}{\sin^2 x},\\ (\sec x)' &= \big(\frac{1}{\cos x}\big)' = \frac{-\cos' x}{\cos^2 x} =\frac{\sin x}{\cos^2 x} = \sec x\tan x, \\ (\csc x)' &= \big(\sec(\frac{\pi}{2} - x)\big)' = -\sec(\frac{\pi}{2} - x)\tan(\frac{\pi}{2} - x) \\ &= -\csc x \cot x \end{align*}

The derivatives of the inverse trigonometric functions are given by \begin{gather*} (\arcsin x)' = \frac{1}{\sqrt{1-x^2}},\quad (\arccos x)' = -\frac{1}{\sqrt{1-x^2}},\\ (\arctan x)' = \frac{1}{1+x^2},\quad (\arccot x)' = -\frac{1}{1+x^2}. \end{gather*}

Proofs: We use the formula for the derivative of the inverse function and the trigonometric identities $$$ \arccos x + \arcsin x =\pi/2,\quad \arctan x + \arccot x =\pi/2. $$$

We get \begin{align*} (\arcsin x)' &= \frac{1}{\sin'(\arcsin x)}= \frac{1}{\cos(\arcsin x)} \\ &= \frac{1}{\sqrt{1-\sin^2(\arcsin x)^2}}= \frac{1}{\sqrt{1-x^2}},\\ (\arccos x)' &= (\pi/2 - \arcsin x)' = -\frac{1}{\sqrt{1-x^2}},\\ (\arctan x)' &= \frac{1}{\tan'(\arctan x)} = \frac{1}{1+\tan^2(\arctan x)} = \frac{1}{1+x^2},\\ (\arccot x)' &= (\pi/2 - \arctan x)' =-\frac{1}{1+x^2}. \end{align*}