Core Techniques — Standard Derivatives, Rules, Chain & Logarithmic

$f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ — the slope of the tangent at $x$. Equivalently $f'(a)=\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$. Geometrically it is the slope of the tangent line; physically, a rate of change.

• Read the definition backwards too: a limit of the shape $\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ or $\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$ IS $f'(a)$ — e.g. $\lim_{x\to\pi/2}\dfrac{\sin x-1}{x-\pi/2}=\cos\tfrac\pi2=0$, and $\lim_{h\to0}\dfrac{f(a+2h)-f(a)}{h}=2f'(a)$.
• From slope to line and angle: the tangent at $(x_1,y_1)$ is $y-y_1=f'(x_1)(x-x_1)$, the normal has slope $-1/f'(x_1)$. The tangent's angle of inclination $\theta\in[0,\pi)$ with the positive x-axis satisfies $\tan\theta=f'(x_1)$ — so a negative slope means an OBTUSE angle, $\theta=\pi-\tan^{-1}|f'|$, and the angle with the y-axis is $\tfrac\pi2-\theta$ (or $\theta-\tfrac\pi2$).

Memorise these; the rules (product, quotient, chain) combine them. Angles are in radians — a degree argument must be converted first.

• Product: $(uv)' = u'v + uv'$.
• Quotient: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
• Linearity: $(au \pm bv)' = au' \pm bv'$.
• A determinant whose entries are functions of $x$ is a sum of products, so the product rule applies row by row: $\dfrac{d}{dx}|R_1;R_2;R_3| = |R_1';R_2;R_3| + |R_1;R_2';R_3| + |R_1;R_2;R_3'|$ (differentiate ONE row at a time, keep the others). For a $2\times2$ or $3\times3$ with simple entries it is usually faster to expand first and then differentiate — full treatment in the Matrices & Determinants notes (linked below).

$\dfrac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$. For nested layers, multiply each layer's derivative: $\dfrac{d}{dx}f(g(h(x))) = f'(g(h(x)))\,g'(h(x))\,h'(x)$.

• Recognisable chain-rule patterns: $u\,u'+v\,v'=\tfrac12\dfrac{d}{dx}(u^2+v^2)$ (so $\sin x\cos x+\dots$ sums often collapse); $\dfrac{d}{dx}\ln|x|=\dfrac1x$ for EVERY $x\ne0$ — for $x<0$, $\ln|x|=\ln(-x)$ and the chain rule gives $\dfrac{-1}{-x}=\dfrac1x$, so the sign does NOT flip; $\dfrac{d}{dx}|x|=\dfrac{x}{|x|}=\operatorname{sgn}x$.
• Parity passes to the derivative, flipped: if $f$ is even then $f'$ is odd, and if $f$ is odd then $f'$ is even (differentiate $f(-x)=\pm f(x)$ by the chain rule: $-f'(-x)=\pm f'(x)$). An odd $f'$ has $f'(0)=0$.

Take $\ln y$, simplify with log laws, then differentiate (the left side gives $\frac{1}{y}\frac{dy}{dx}$):

• Variable exponent: $y = f(x)^{g(x)} \Rightarrow \ln y = g(x)\ln f(x)$.
• Product of powers: $\ln y$ splits into a sum, each term differentiated alone.
• Power tower: $y = (f(x))^{y} \Rightarrow \ln y = y\ln f(x)$ (the exponent is the whole $y$ again).

To find $\dfrac{du}{dv}$ where $u=u(x)$ and $v=v(x)$: compute $\dfrac{du}{dx}$ and $\dfrac{dv}{dx}$, then $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$. It is the same idea as parametric differentiation, with $x$ as the hidden parameter.

Common collapses (memorise the substitutions):

• $\tan^{-1}\!\dfrac{2x}{1-x^2},\ \sin^{-1}\!\dfrac{2x}{1+x^2},\ \cos^{-1}\!\dfrac{1-x^2}{1+x^2}$: put $x=\tan\theta\Rightarrow 2\theta = 2\tan^{-1}x$.
• $\cos^{-1}(\sin x) = \tfrac{\pi}{2}-x$; $\tan^{-1}\!\dfrac{a-b}{1+ab}=\tan^{-1}a-\tan^{-1}b$.
• Complementary pairs are constants, so their derivative is 0: $\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\sec^{-1}x+\csc^{-1}x=\tfrac\pi2$. Hence $\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^2}}$ and $\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}$ — the co-function's derivative is the negative of its partner's.
• Reciprocal arguments: $\sec^{-1}z=\cos^{-1}\tfrac1z$, $\csc^{-1}z=\sin^{-1}\tfrac1z$, $\cot^{-1}z=\tan^{-1}\tfrac1z$ (for $z>0$); so $y=\sec^{-1}\!\big(\tfrac{1}{2x^2-1}\big)=\cos^{-1}(2x^2-1)=2\cos^{-1}x$ on $[0,1]$.
• Un-collapsing on the wrong branch: $\cos^{-1}(\cos x)=x$ only on $[0,\pi]$; on $[-\pi,0]$ it is $-x$, so $\cos^{-1}(\cos x)=|x|$ on $[-\pi,\pi]$ and its derivative is $-1$ for $x<0$. Likewise $\sin^{-1}(\sin x)=\pi-x$ on $[\tfrac\pi2,\tfrac{3\pi}2]$, derivative $-1$.

Differentiate the collapsed form (often $\pm 1$, $\pm 2/(1+x^2)$, etc.).

For $f(x+y)=f(x)f(y)$: from first principles $f'(x)=f(x)\lim_{h\to 0}\dfrac{f(h)-1}{h}=f'(0)\,f(x)$. More generally, differentiate the given relation w.r.t. one variable and substitute convenient values (often $y=0$) to expose $f'$.

Keep these collapses in reflex memory:

• Half-angle of $1\pm\cos x$: $1-\cos x = 2\sin^2\tfrac{x}{2}$, $1+\cos x = 2\cos^2\tfrac{x}{2}$; so $\dfrac{1-\cos x}{1+\cos x}=\tan^2\tfrac{x}{2}$, $\dfrac{1}{1-\cos x}=\tfrac12\csc^2\tfrac{x}{2}$, $\dfrac{1}{1+\cos x}=\tfrac12\sec^2\tfrac{x}{2}$.
• Power-reduction (double angle): $1-\cos 2x = 2\sin^2 x$, $1+\cos 2x = 2\cos^2 x$, $\sin 2x = 2\sin x\cos x$.
• Perfect square under a root: $1\pm\sin 2x=(\sin x\pm\cos x)^2$ and $1\pm\sin\theta=\left(\cos\tfrac\theta2\pm\sin\tfrac\theta2\right)^2$, so $\sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|$ — keep the modulus; its sign depends on the interval.
• $\sec\pm\tan$: $\sec x+\tan x=\dfrac{1+\sin x}{\cos x}=\tan\!\left(\tfrac\pi4+\tfrac x2\right)$, $\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)$.
• Harmonic form: $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\alpha)$, so its extreme values are $\pm\sqrt{a^2+b^2}$.
• Weierstrass $t=\tan\tfrac{x}{2}$: $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ — useful whenever a rational function of $\sin x,\cos x$ must be handled in one variable.
• The elementary ones, which the exotic collapses assume: $\sin^2x+\cos^2x=1$, $1+\tan^2x=\sec^2x$, $1+\cot^2x=\csc^2x$. They are how $\sec^2(\tan^{-1}x)=1+x^2$ and $\cos(\tan^{-1}x)=\dfrac{1}{\sqrt{1+x^2}}$ are read off.
• Algebraic collapses before differentiating (take logs of a product only AFTER simplifying it): $(1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{k-1}})=1-x^{2^k}$ (telescoping difference of squares); $1+x^4+x^8=(1+x^2+x^4)(1-x^2+x^4)$, from $a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)$; $x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$.