JEE Mathematics Methods of Differentiation Notes Set 02

Meaning of the Derivative and the Basic Rules

A function \( y=f(x) \) is differentiable at \( a \) when the limit \( \lim_{h\to0}\frac{f(a+h)-f(a)}{h} \) exists and is finite. The value of the limit is written \( f'(a) \) or \( \left(\frac{dy}{dx}\right)_{x=a} \).
Linear combination: \( (c_1f\pm c_2g)'=c_1f'\pm c_2g' \). Product: \( (fg)'=fg'+gf' \), and for three factors each factor is differentiated in turn while the other two are kept. Quotient: \( \left(\frac fg\right)'=\frac{gf'-fg'}{g^2} \).
Chain rule: if \( y=f(t) \) and \( t=g(x) \) then \( \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx} \). For three or more links, multiply the derivatives of every link from the outside in.

Standard Derivatives

Algebraic and exponential: \( (x^n)'=nx^{n-1} \), \( (e^x)'=e^x \), \( (a^x)'=a^x\ln a \), \( (\sqrt x)'=\frac{1}{2\sqrt x} \), \( (|x|)'=\frac{|x|}{x} \) for \( x\ne0 \), \( (\ln|x|)'=\frac1x \) and \( (\log_a|x|)'=\frac{1}{x\ln a} \). The greatest integer function \( [x] \) has derivative \( 0 \) except at integers, where it is not differentiable.
Trigonometric: \( (\sin x)'=\cos x \), \( (\cos x)'=-\sin x \), \( (\tan x)'=\sec^2x \), \( (\cot x)'=-\mathrm{cosec}^2x \), \( (\sec x)'=\sec x\tan x \), \( (\mathrm{cosec}\,x)'=-\mathrm{cosec}\,x\cot x \).
Inverse trigonometric: \( (\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}} \), \( (\cos^{-1}x)'=\frac{-1}{\sqrt{1-x^2}} \), \( (\tan^{-1}x)'=\frac{1}{1+x^2} \), \( (\cot^{-1}x)'=\frac{-1}{1+x^2} \), \( (\sec^{-1}x)'=\frac{1}{|x|\sqrt{x^2-1}} \), \( (\mathrm{cosec}^{-1}x)'=\frac{-1}{|x|\sqrt{x^2-1}} \).
Hyperbolic: \( (\sinh x)'=\cosh x \), \( (\cosh x)'=\sinh x \), \( (\tanh x)'=\mathrm{sech}^2x \), \( (\coth x)'=-\mathrm{cosech}^2x \), \( (\mathrm{sech}\,x)'=-\mathrm{sech}\,x\tanh x \), \( (\mathrm{cosech}\,x)'=-\mathrm{cosech}\,x\coth x \).
Inverse hyperbolic: \( (\sinh^{-1}x)'=\frac{1}{\sqrt{x^2+1}} \), \( (\cosh^{-1}x)'=\frac{1}{\sqrt{x^2-1}} \) for \( x>1 \), \( (\tanh^{-1}x)'=\frac{1}{1-x^2} \) for \( |x|<1 \), \( (\coth^{-1}x)'=\frac{1}{1-x^2} \) for \( |x|>1 \), \( (\mathrm{sech}^{-1}x)'=\frac{-1}{x\sqrt{1-x^2}} \) for \( 0<x<1 \), \( (\mathrm{cosech}^{-1}x)'=\frac{-1}{|x|\sqrt{1+x^2}} \).

Useful Forms for Quick Differentiation

If \( y=\ln\left|f(x)\right| \) then \( y'=\frac{f'(x)}{f(x)} \), and if \( y=\sqrt{f(x)} \) then \( y'=\frac{f'(x)}{2\sqrt{f(x)}} \). For a Mobius function, \( \frac{d}{dx}\left(\frac{ax+b}{cx+d}\right)=\frac{ad-bc}{(cx+d)^2} \).
If \( f(x+y)=f(x)f(y) \) for all real \( x,y \) and \( f(x)\ne0 \) then \( f'(x)=f'(0)f(x) \).
Infinite radicals and exponent towers satisfy a self-referencing relation. For \( y=\sqrt{f(x)+\sqrt{f(x)+\cdots}} \) we have \( y^2=f(x)+y \), so \( y'=\frac{f'(x)}{2y-1} \). For \( y=\sqrt[n]{f(x)+\sqrt[n]{f(x)+\cdots}} \) we have \( y^n=f(x)+y \), so \( y'=\frac{f'(x)}{ny^{n-1}-1} \). If \( y=f(x)+\frac1y \) then \( y'=\frac{y^2f'(x)}{y^2+1} \).

Choosing a Substitution Before Differentiating

When an expression contains \( \sqrt{a^2-x^2} \) put \( x=a\sin\theta \) or \( x=a\cos\theta \). For \( \sqrt{a^2+x^2} \) put \( x=a\tan\theta \). For \( \sqrt{x^2-a^2} \) put \( x=a\sec\theta \). For \( \sqrt{\frac{a-x}{a+x}} \) or \( \sqrt{\frac{a+x}{a-x}} \) put \( x=a\cos2\theta \). For \( \sqrt{(a-x)(x-b)} \) put \( x=a\cos^2\theta+b\sin^2\theta \). For \( \sqrt{2ax-x^2} \) put \( x=2a\sin^2\theta \).
After the substitution, use a trigonometric identity to reduce the inverse trigonometric function to a multiple of the angle, for example \( \tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x \) for \( |x|<1 \) and \( \tan^{-1}\frac{3x-x^3}{1-3x^2}=3\tan^{-1}x \). Check the interval, because the principal value of the inverse function decides whether the answer carries a plus or minus sign.

Implicit and Parametric Differentiation

Implicit: for \( F(x,y)=0 \) differentiate every term with respect to \( x \), treat \( y \) as a function of \( x \), and collect the \( \frac{dy}{dx} \) terms. Equivalently, \( \frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y} \).
For the general conic \( ax^2+2hxy+by^2+2gx+2fy+c=0 \) this gives \( \frac{dy}{dx}=-\frac{ax+hy+g}{hx+by+f} \).
Shortcuts: if \( f\left(\frac yx\right)=C \) then \( \frac{dy}{dx}=\frac yx \); if \( f(xy)=C \) then \( \frac{dy}{dx}=-\frac yx \); if \( f(x^my^n)=C \) then \( \frac{dy}{dx}=-\frac{my}{nx} \). A homogeneous equation in \( x \) and \( y \) also gives \( \frac{dy}{dx}=\frac yx \).
Parametric: for \( x=f(t),\ y=g(t) \), \( \frac{dy}{dx}=\frac{g'(t)}{f'(t)} \) and \( \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{1}{f'(t)} \).

Logarithmic Differentiation

Use it for products, quotients, and powers with variable exponents. If \( y=f_1(x)^{f_2(x)} \) take logarithms to get \( \ln y=f_2\ln f_1 \) and differentiate. For \( y=f(x)^{g(x)} \) the result is \( y'=f(x)^{g(x)}\left[g(x)\frac{f'(x)}{f(x)}+g'(x)\ln f(x)\right] \).
Two useful results: \( \frac{d}{dx}\left(x^x\right)=x^x(1+\ln x) \), and if \( y=f(x)^y \) then \( \frac{dy}{dx}=\frac{y^2f'(x)}{f(x)\left[1-y\ln f(x)\right]} \).

Derivative of a Determinant, Inverse and Composite Functions

If the entries of a determinant are functions of \( x \), its derivative is the sum of determinants obtained by differentiating one row (or one column) at a time and keeping the other rows unchanged.
If \( g \) is the inverse of \( f \) then \( g'(f(x))=\frac{1}{f'(x)} \) and \( g''(f(x))=-\frac{f''(x)}{(f'(x))^3} \). For a composite function \( h=f\circ g \), \( h'(x)=f'(g(x))g'(x) \), so \( \frac{h'(x)}{h(x)} \) is the logarithmic derivative of \( f \) evaluated at \( g(x) \). The derivative of \( u \) with respect to \( v \) is \( \frac{du/dx}{dv/dx} \).

Higher Order Derivatives and Properties

\( \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right) \), and in the same way for higher orders. The derivative of an even function is odd and the derivative of an odd function is even. If \( f \) is even and differentiable then \( f'(0)=0 \). If \( f \) is differentiable and periodic then \( f' \) is periodic with the same period.
If a differentiable graph is symmetric about the vertical line \( x=a \) then \( f'(a)=0 \). A modulus function \( |g(x)| \) is not differentiable at a point where \( g \) crosses zero with a non-zero slope. If \( |f(x)-f(y)|\le|x-y|^p \) for some \( p>1 \), then \( f \) is constant.

Worked Examples

Example 1. Find \( \frac{d}{dx}\left[(x+1)(x^2+1)(x^4+1)(x^8+1)\right] \).
Answer:
The product telescopes: multiply and divide by \( x-1 \) to get \( \frac{x^{16}-1}{x-1} \).
Differentiating by the quotient rule: \( \frac{16x^{15}(x-1)-(x^{16}-1)}{(x-1)^2}=\frac{15x^{16}-16x^{15}+1}{(x-1)^2} \).

Example 2. If \( y=\tan^{-1}\sqrt{\frac{1-x}{1+x}} \), find \( \frac{dy}{d(\cos^{-1}x)} \).
Answer:
Put \( x=\cos2\theta \). Then \( y=\tan^{-1}(\tan\theta)=\theta=\frac12\cos^{-1}x \).
So \( \frac{dy}{d(\cos^{-1}x)}=\frac12 \).

Example 3. If \( x=a\cos^3\theta \) and \( y=a\sin^3\theta \), find \( \frac{dy}{dx} \).
Answer:
\( \frac{dx}{d\theta}=-3a\cos^2\theta\sin\theta \) and \( \frac{dy}{d\theta}=3a\sin^2\theta\cos\theta \).
Dividing gives \( \frac{dy}{dx}=-\tan\theta \).

Example 4. If \( x^y=y^x \), find \( \frac{dy}{dx} \).
Answer:
Take logarithms: \( y\ln x=x\ln y \). Differentiate: \( y'\ln x+\frac yx=\ln y+\frac xyy' \).
Collect the terms with \( y' \): \( y'=\frac{y(x\ln y-y)}{x(y\ln x-x)} \).

Example 5. If \( y=\sin(\ln x) \), show that \( x^2y''+xy'+y=0 \).
Answer:
\( y'=\frac{\cos(\ln x)}{x} \), so \( xy'=\cos(\ln x) \).
Differentiating again: \( xy''+y'=-\frac{\sin(\ln x)}{x} \). Multiply by \( x \): \( x^2y''+xy'=-y \), which is the required result.

Example 6. Transform \( (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=0 \) by putting \( x=\cos t \).
Answer:
\( \frac{dy}{dx}=-\frac{1}{\sin t}\frac{dy}{dt} \) and \( \frac{d^2y}{dx^2}=\frac{1}{\sin^2t}\frac{d^2y}{dt^2}-\frac{\cos t}{\sin^3t}\frac{dy}{dt} \).
Substituting with \( 1-x^2=\sin^2t \), the first-order terms cancel and the equation becomes \( \frac{d^2y}{dt^2}+y=0 \).

Example 7. Sum the series \( 1+2x+3x^2+\cdots+nx^{n-1} \) for \( x\ne1 \).
Answer:
Start from \( 1+x+\cdots+x^n=\frac{x^{n+1}-1}{x-1} \) and differentiate both sides.
The sum is \( \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} \).

Example 8. If \( f(x+y)=f(x)f(y) \), \( f(5)=2 \) and \( f'(0)=3 \), find \( f'(5) \).
Answer:
From the functional equation, \( f'(x)=f(x)f'(0) \).
So \( f'(5)=2\times3=6 \).