Derivatives of basic elementary functions

With the general rules of §10.4 in hand, we now derive — once, from first principle, each time using the $h$-form $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ — the derivatives of every basic elementary function. Once this table exists, no later problem in the chapter needs to return to a first-principle limit.

(1) Constant function. For $f(x)=k$ ($k$ constant), $f(x+h)-f(x)=k-k=0$ for every $h$, so the difference quotient is identically $0$ and $\dfrac{d}{dx}(k)=0$.

(2) Power function, $y=x^n$ ($n$ a positive integer). $f(x+h)-f(x)=(x+h)^n-x^n$. Writing $y=x+h$ and factoring the difference of $n$th powers, $\dfrac{(x+h)^n-x^n}{h} = \dfrac{y^n-x^n}{y-x}\to nx^{n-1}$ as $h\to0$ (equivalently $y\to x$), so $\dfrac{d}{dx}(x^n)=nx^{n-1}$.

Corollary 10.1. When $n=p/q$ (a rational exponent, $p,q$ integers, $q\ne0$), the same power formula holds: $\dfrac{d}{dx}\!\left(x^{p/q}\right) = \dfrac{p}{q}\,x^{\frac{p}{q}-1}$.

Corollary 10.2. For any real number $\alpha$, $\dfrac{d}{dx}(x^\alpha) = \alpha\,x^{\alpha-1}$ — the power rule holds for every real exponent, not just integers.

For instance: $\dfrac{d}{dx}(5)=0$ (a constant); $\dfrac{d}{dx}(x^3)=3x^2$; $\dfrac{d}{dx}\!\left(x^{3/2}\right)=\tfrac32 x^{1/2}$; $\dfrac{d}{dx}(x^2)=2x$; $\dfrac{d}{dx}\!\left(x^{2/3}\right)=\tfrac23 x^{-1/3}$ ($x\ne0$); and, by the constant-multiple rule, $\dfrac{d}{dx}(100x^9) = 100\cdot 9x^8 = 900x^8$.

(3) Logarithmic function. For $f(x)=\log x$ (natural log, base $e$), $f(x+h)-f(x)=\log\!\left(\dfrac{x+h}{x}\right) = \log\!\left(1+\dfrac{h}{x}\right)$, so $\dfrac{f(x+h)-f(x)}{h} = \dfrac1x\cdot\dfrac{\log(1+h/x)}{h/x}$. Using the standard limit $\lim_{\alpha\to0}\dfrac{\log(1+\alpha)}{\alpha}=1$ (with $\alpha=h/x\to0$) gives $\dfrac{d}{dx}(\log x)=\dfrac1x$.

Corollary 10.3. For $y=\log_a x = \dfrac{\log x}{\log a}$ (change of base, $\log x$ = natural log), the constant-multiple rule gives $\dfrac{d}{dx}(\log_a x) = \dfrac{1}{x\log a}$.

(4) Exponential function. For $f(x)=a^x$ ($a>0$), $\dfrac{f(x+h)-f(x)}{h} = a^x\left(\dfrac{a^h-1}{h}\right)$, and the standard limit $\lim_{h\to0}\dfrac{a^h-1}{h}=\log a$ gives $\dfrac{d}{dx}(a^x) = a^x\log a$. In the special case $a=e$ (so $\log e = 1$), $\dfrac{d}{dx}(e^x)=e^x$ — the exponential function is its own derivative.

(5) The six trigonometric functions.

• Sine. $f(x+h)-f(x) = \sin(x+h)-\sin x = 2\sin\!\left(\tfrac h2\right)\cos\!\left(x+\tfrac h2\right)$ (sum-to-product identity), so $\dfrac{f(x+h)-f(x)}{h} = \dfrac{\sin(h/2)}{h/2}\cdot\cos\!\left(x+\tfrac h2\right) \to 1\cdot\cos x$ as $h\to0$ (using $\lim_{\theta\to0}\sin\theta/\theta=1$ and continuity of $\cos$). So $\dfrac{d}{dx}(\sin x)=\cos x$.
• Cosine. Using $\cos x = \sin\!\left(\tfrac\pi2+x\right)$ and the chain rule with inner function $u=\tfrac\pi2+x$ ($u'=1$): $\dfrac{d}{dx}(\cos x) = \cos u \cdot 1 = \cos\!\left(\tfrac\pi2+x\right) = -\sin x$. So $\dfrac{d}{dx}(\cos x)=-\sin x$.
• Tangent. $\tan x = \sin x/\cos x$; the quotient rule gives $\dfrac{d}{dx}(\tan x) = \dfrac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2x} = \dfrac{\cos^2x+\sin^2x}{\cos^2x}=\dfrac{1}{\cos^2x}=\sec^2 x$.
• Secant. $\sec x = (\cos x)^{-1}$; the chain rule gives $\dfrac{d}{dx}(\sec x) = -(\cos x)^{-2}(-\sin x) = \dfrac{\sin x}{\cos^2x} = \sec x\tan x$.
• Cosecant. $\operatorname{cosec}x = (\sin x)^{-1}$; similarly $\dfrac{d}{dx}(\operatorname{cosec}x) = -(\sin x)^{-2}(\cos x) = -\operatorname{cosec}x\cot x$.
• Cotangent. $\cot x = \cos x/\sin x$; the quotient rule gives $\dfrac{d}{dx}(\cot x) = \dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x} = \dfrac{-(\sin^2x+\cos^2x)}{\sin^2x} = -\operatorname{cosec}^2x$.

(6) The six inverse trigonometric functions.

• $\sin^{-1}x$. Let $y=\sin^{-1}x$, so $x=\sin y$; differentiating both increments and passing to the limit as in implicit differentiation (§10.4.3) gives $\dfrac{dy}{dx}=\dfrac{1}{\cos y}=\dfrac{1}{\sqrt{1-\sin^2y}}=\dfrac{1}{\sqrt{1-x^2}}$ (taking the principal branch, $\cos y\ge0$). So $\dfrac{d}{dx}\!\left(\sin^{-1}x\right)=\dfrac{1}{\sqrt{1-x^2}}$.
• $\cos^{-1}x$. From the identity $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ (a constant), differentiating both sides gives $\dfrac{d}{dx}\!\left(\cos^{-1}x\right) = -\dfrac{d}{dx}\!\left(\sin^{-1}x\right) = -\dfrac{1}{\sqrt{1-x^2}}$.
• $\tan^{-1}x$. Let $y=\tan^{-1}x$, so $x=\tan y$; the same implicit-style limit argument gives $\dfrac{dy}{dx}=\dfrac{1}{\sec^2y}=\dfrac{1}{1+\tan^2y}=\dfrac{1}{1+x^2}$. So $\dfrac{d}{dx}\!\left(\tan^{-1}x\right)=\dfrac{1}{1+x^2}$. …