JEE Mathematics Differentiation Notes Set 01

Mathematics Concept Notes for JEE: Differentiation

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Meaning of the Derivative

The derivative of \( f \) at a point \( a \) is the limit \( f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h} \), when it exists. Geometrically it is the slope of the tangent to the graph at \( x=a \); physically it is the instantaneous rate of change, for example the speed of a particle when f gives its distance. A function can be differentiated at a point only if it is continuous there and its left and right derivatives are equal. Moduli, greatest-integer functions and piecewise definitions need this check at the joining points.

The symbol \( \dfrac{dy}{dx} \) is a single operator applied to y, not a fraction of two separate quantities. If \( f'(a) \) is infinite, the tangent at that point is vertical.

Standard Derivatives

FunctionDerivative
\( x^n \)\( nx^{n-1} \)
\( \sin x,\ \cos x \)\( \cos x,\ -\sin x \)
\( \tan x,\ \cot x \)\( \sec^2x,\ -\csc^2x \)
\( \sec x,\ \csc x \)\( \sec x\tan x,\ -\csc x\cot x \)
\( \ln x,\ \log_ax \)\( \dfrac1x,\ \dfrac1{x\ln a} \)
\( e^x,\ a^x \)\( e^x,\ a^x\ln a \)
\( \sin^{-1}x,\ \cos^{-1}x \)\( \dfrac1{\sqrt{1-x^2}},\ -\dfrac1{\sqrt{1-x^2}} \)
\( \tan^{-1}x,\ \cot^{-1}x \)\( \dfrac1{1+x^2},\ -\dfrac1{1+x^2} \)
\( \sec^{-1}x,\ \csc^{-1}x \)\( \dfrac{1}{|x|\sqrt{x^2-1}},\ -\dfrac1{|x|\sqrt{x^2-1}} \)
\( \sinh x,\ \cosh x,\ \tanh x \)\( \cosh x,\ \sinh x,\ \operatorname{sech}^2x \)
\( \sinh^{-1}x,\ \cosh^{-1}x \)\( \dfrac1{\sqrt{x^2+1}},\ \dfrac1{\sqrt{x^2-1}} \)

Rules of Differentiation

1. Sum, difference and constant multiple: \( (cf\pm g)'=cf'\pm g' \).
2. Product rule: \( (fg)'=f'g+fg' \); for three factors differentiate one factor at a time.
3. Quotient rule: \( \left(\dfrac fg\right)'=\dfrac{f'g-fg'}{g^2} \), provided \( g\ne0 \).
4. Chain rule: \( \dfrac d{dx}f(g(x))=f'(g(x))\,g'(x) \). For three nested functions multiply three derivatives, working from the outside in.
5. Reciprocal relation: \( \dfrac{dy}{dx}=\dfrac1{dx/dy} \), useful when x is given as a function of y or for inverse functions: if g is the inverse of f then \( g'(x)=\dfrac1{f'(g(x))} \).
6. If \( f'(x)=0 \) on an interval, f is constant there.

Simplify Before You Differentiate

Many inverse trigonometric expressions collapse to a simple function once a suitable substitution is made, so differentiation becomes trivial. Useful identities are \( \sin^{-1}x+\cos^{-1}x=\frac\pi2 \), \( \tan^{-1}x+\cot^{-1}x=\frac\pi2 \), \( \tan^{-1}x\pm\tan^{-1}y=\tan^{-1}\frac{x\pm y}{1\mp xy} \), \( 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}=\sin^{-1}\frac{2x}{1+x^2}=\cos^{-1}\frac{1-x^2}{1+x^2} \), \( 3\sin^{-1}x=\sin^{-1}(3x-4x^3) \) and \( 3\cos^{-1}x=\cos^{-1}(4x^3-3x) \). Each identity is valid only on a restricted range of x, so check the principal range of the inverse function before using it.

Expression in the questionSubstitution that simplifies it
\( \sqrt{a^2-x^2} \)\( x=a\sin\theta \) or \( a\cos\theta \)
\( \sqrt{a^2+x^2} \)\( x=a\tan\theta \)
\( \sqrt{x^2-a^2} \)\( x=a\sec\theta \)
\( \sqrt{\dfrac{a-x}{a+x}} \)\( x=a\cos2\theta \)
\( \sqrt{\dfrac{a^2-x^2}{a^2+x^2}} \)\( x^2=a^2\cos2\theta \)
\( \sqrt{ax-x^2} \)\( x=a\sin^2\theta \)

Implicit, Logarithmic and Parametric Differentiation

1. Implicit functions: differentiate every term with respect to x, treating y as a function of x, then collect the terms containing \( \frac{dy}{dx} \). For \( f(x,y)=c \), \( \dfrac{dy}{dx}=-\dfrac{\partial f/\partial x}{\partial f/\partial y} \).
2. Logarithmic differentiation: for \( y=[f(x)]^{g(x)} \) or for long products and quotients take logarithms first. For the power form \( y'=y\left[g'\ln f+\dfrac{gf'}f\right] \).
3. Parametric equations \( x=\phi(t),\ y=\psi(t) \): \( \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt} \). Eliminate the parameter only if that is easy.
4. Derivative of f(x) with respect to g(x): divide the two derivatives, \( \dfrac{df}{dg}=\dfrac{df/dx}{dg/dx} \).

Infinite Nested Expressions

When a quantity repeats itself inside, replace the tail by y and differentiate the resulting finite relation. For \( y=\sqrt{f(x)+\sqrt{f(x)+\cdots}} \) we get \( y^2=f+y \), so \( y'=\dfrac{f'}{2y-1} \). For the power tower \( y=f^{f^{f\cdots}} \) we get \( y=f^y \) and \( y'=\dfrac{y^2f'}{f(1-y\ln f)} \). For the continued fraction \( y=f+\dfrac1{f+\cdots} \) we get \( y=f+\frac1y \) and \( y'=\dfrac{yf'}{2y-f} \).

Successive Differentiation

The second derivative is the derivative of the first, and so on; the n-th derivative is written \( \dfrac{d^ny}{dx^n},\ y_n,\ f^{(n)}(x) \) or \( D^ny \). The standard n-th derivatives are collected below.

Functionn-th derivative
\( (ax+b)^m \), with \( m>n \)\( \dfrac{m!}{(m-n)!}a^n(ax+b)^{m-n} \)
\( x^n \)\( n! \)
\( \dfrac1{ax+b} \)\( \dfrac{(-1)^nn!\,a^n}{(ax+b)^{n+1}} \)
\( \ln(ax+b) \)\( \dfrac{(-1)^{n-1}(n-1)!\,a^n}{(ax+b)^n} \)
\( e^{ax},\ a^x \)\( a^ne^{ax},\ a^x(\ln a)^n \)
\( \sin(ax+b),\ \cos(ax+b) \)\( a^n\sin\!\left(ax+b+\frac{n\pi}2\right),\ a^n\cos\!\left(ax+b+\frac{n\pi}2\right) \)
\( e^{ax}\sin(bx+c) \)\( r^ne^{ax}\sin(bx+c+n\phi) \), with \( r=\sqrt{a^2+b^2},\ \tan\phi=\frac ba \)

For a rational function, first divide (if the numerator has the higher degree) and then split into partial fractions; each partial fraction \( \frac1{x-p} \) has the n-th derivative given above, and the polynomial part disappears after enough differentiations.

Leibnitz Theorem

For a product of two functions, \( D^n(uv)=\sum_{r=0}^{n}{}^nC_r\,D^{n-r}u\,D^rv \). Choose v as the factor that becomes zero after a few differentiations, for example a polynomial, so that only a few terms of the sum survive. Example: for \( y=x^2e^x \), taking \( v=x^2 \) gives \( y_n=e^x\left[x^2+2nx+n(n-1)\right] \).

Differentiating Determinants and Integrals

1. Determinant: differentiate one row (or column) at a time keeping the others unchanged and add the resulting determinants.
2. Integral with variable limits: \( \dfrac d{dx}\int_{g_1(x)}^{g_2(x)}f(t)\,dt=f(g_2(x))\,g_2'(x)-f(g_1(x))\,g_1'(x) \).
3. Functional equations such as \( f(x+y)=f(x)f(y) \) give \( f'(x)=f(x)f'(0) \); always find \( f(0) \) first.

Worked Examples

1. Differentiate \( y=\tan^{-1}\dfrac{2x}{1-x^2} \) for \( |x|<1 \). Put \( x=\tan\theta \): \( y=2\theta=2\tan^{-1}x \), so \( y'=\dfrac2{1+x^2} \).
2. Find \( \dfrac{dy}{dx} \) for \( y=x^x \). Taking logs, \( \ln y=x\ln x \), so \( \dfrac{y'}y=\ln x+1 \) and \( y'=x^x(1+\ln x) \).
3. Find \( \dfrac{dy}{dx} \) if \( x=a(\theta-\sin\theta),\ y=a(1-\cos\theta) \). \( \dfrac{dy}{dx}=\dfrac{a\sin\theta}{a(1-\cos\theta)}=\cot\dfrac\theta2 \).
4. If \( \sin y=x\sin(a+y) \), differentiate implicitly: \( \cos y\,y'=\sin(a+y)+x\cos(a+y)\,y' \), so \( y'=\dfrac{\sin(a+y)}{\cos y-x\cos(a+y)}=\dfrac{\sin^2(a+y)}{\sin a} \).

Revision Notes and Key Concepts for JEE Mathematics Differentiation

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