Differentiation

Differentiation is a mathematical process that measures how a function changes as its input changes. In simpler terms, it calculates the rate of change of function. 

The derivative can be visualized as the slope of the tangent line to the functions graphs at the point.

Differentiation is used to calculate rates of change in various field, such as physics (velocity, acceleration…..) 

1.0Derivative By First Principle

Let y = f(x) : 

y + Δy = f(x + Δx)

$\therefore \frac{\Delta \mathrm{y}}{\Delta \mathrm{x}}=\frac{\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})}{\Delta \mathrm{x}}$(average rate of change of function)

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}={\lim }_{\Delta \mathrm{x}\to 0}\frac{\Delta \mathrm{y}}{\Delta \mathrm{x}}={\lim }_{\Delta \mathrm{x}\to 0}\frac{\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})}{\Delta \mathrm{x}}$ …..(i)

(i) denotes the instantaneous rate of change of function and gives slope of tangent at any point on curve y = f(x). Finding the value of the limit given by (i) in respect to a variety of functions is called finding the derivative by first principle/by delta method/by ab-initio method/by fundamental definition of calculus.

Note than if y = f(x) then the symbols $\frac{dy}{dx}=Dy=f^{\prime }(x)=y_1$ or y’ have the same meaning.

Derivative of standard functions:

2.0Fundamental Theorems

If f and g are derivable functions of x, then,

(a) $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}\pm \mathrm{g})=\frac{\mathrm{df}}{\mathrm{dx}}\pm \frac{\mathrm{dg}}{\mathrm{dx}}$

(b) $\frac{d}{dx}(cf)=c\frac{df}{dx}$, where c is any constant

(c) $\frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}$ known as “PRODUCT RULE”

(d) $\frac{d}{dx}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)-\mathrm{f}\left(\frac{\mathrm{dg}}{\mathrm{dx}}\right)}{{\mathrm{g}}^2}$ where g ≠ 0 known as “QUOTIENT RULE”

(e) If y = f(u) & u = g (x) then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ known as “CHAIN RULE”

Note : in general if y = f(u) then $\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{f}}^{\prime }(\mathrm{u})\cdot \frac{\mathrm{du}}{\mathrm{dx}}$

1. Logarithmic differentiation:

To find the derivative of a function:

(a) which is the product of quotient of a number of functions or

(b) of the form [f(x)]^g(x) where f and g are both derivable functions.

It is convenient to take the logarithm of the function first and then differentiate.

2. Parametric differentiation:

• If y = f(θ) and x = g(θ) where θ is a parameter, then $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{f^{\prime }(\theta )}{g^{\prime }(\theta )}$

3. Derivative of a function w.r.t. another function

• Let y = f(x) ; z = g(x) then $\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{f^{\prime }(x)}{g^{\prime }(x)}$

4. Differentiation of implicit functions: φ(x, y) = 0

• To find $\frac{dy}{dx}$ of implicit functions, we differentiate each term w.r.t. regarding y as a function of x and then collect terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ together on one side.
• Also $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{\partial \varphi }{\partial x}}{\frac{\partial \varphi }{\partial y}}$, where $\frac{\partial \varphi }{\partial y}$ = partial derivative of φ(x, y) w.r.t. x taking y as a constant and $\frac{\partial \varphi }{\partial y}$= partial derivative of φ(x, y) w.r.t. y taking x as a constant. (In the case of implicit functions, generally, both x and y present in answers of dy/dx)

3.0Derivative Of Inverse Function

Theorem: If the inverse functions f and g are defined by y = f(x) and x = g(y) and if f'(x) exists and f’(x) ≠ 0, then $g’(y)=\frac{1}{f^{\prime }(x)}$. This result can also be written as, if $\frac{\mathrm{dy}}{\mathrm{dx}}$ exist and $\frac{dy}{dx}\ne 0$, then $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$ or $\frac{dy}{dx}\cdot \frac{dx}{dy}=1$ or $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}}\left[\frac{\mathrm{dx}}{\mathrm{dy}}\ne 0\right]$

Note: (i) $g’(f(x))=\frac{1}{f^{\prime }(x)}$ (ii) $g^{\prime \prime }(f(x))=\frac{-f^{\prime \prime }(x)}{{\left(f^{\prime }(x)\right)}^3}$

Higher order Derivatives

Let a function y = f(x) be defined on an interval (a, b). If f(x) is differentiable function, then its derivative f’(x) [or (dy/dx) or y’] is called the first derivative of y w.r.t. x if f’(x) is again differentiable function on (a, b), then its derivative f”(x) [or d²y/dx² or y”] is called second derivative of y w.r.t. x similarly, the 3^rd order derivative of y w.r.t. x, if exists, is defined by $\frac{{\mathrm{d}}^3\mathrm{y}}{{\mathrm{dx}}^3}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)$ and denoted by f”’ (x) or y”’ and so on.

Note: If x = f (θ) and y = g(θ) where ‘θ’ is a parameter then $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\&\frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)/\frac{dx}{d\theta }$, In general $\frac{d^{n}y}{d{x}^n}=\frac{d}{d\theta }\left(\frac{d^{n-1}y}{d{x}^{n-1}}\right)/\frac{dx}{d\theta }$

L. Hopital’s Rule

(a) This rule is applicable for the indeterminate forms of the type $\frac{0}{0},\frac{\infty }{\infty }$. If the function f(x) and g(x) are differentiable in certain neighbourhood of the point ‘a’, except, may be, at the point ‘a’ itself and g'(x) ≠ 0, and if

${\lim }_{x\to a}f(x)={\lim }_{x\to a}g(x)=0$

or ${\lim }_{x\to a}f(x)={\lim }_{x\to a}g(x)=\infty$,

then ${\lim }_{x\to a}\frac{f(x)}{g(x)}={\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

provided the limit ${\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ exists (L’ Hopital’s rule). The point ‘a’ may be either finite or improper ($+\infty \text{ or }-\infty$) 

(b) Indeterminate forms of the type 0. ∞ or ∞ – ∞ are reduced to forms of the type $\frac{0}{0}$ or $\frac{\infty }{\infty }$ by algebraic transformations.

(c) Indeterminate forms of the type 1^∞, ∞⁰ or 0⁰ are reduced to forms of the type 0 × ∞ by taking logarithms or by the transformation $[\mathrm{f}(\mathrm{x}){]}^{\varphi (\mathrm{x})}={\mathrm{e}}^{\varphi (\mathrm{x})\cdot \ln f(\mathrm{x})}$

4.0Solved Examples on Differentiation

Ex.1 Differentiate each of following functions by first principle:

(i) f(x) = tanx (ii) f(x) = e^sinx

Sol.

(i) $f^{\prime }(x)={\lim }_{h\to 0}\frac{\tan (x+h)-\tan x}{h}={\lim }_{h\to 0}\frac{\tan (x+h-x)[1+\tan x\tan (x+h)]}{h}$

$=f^{\prime }(x)={\lim }_{h\to 0}\frac{\tan h}{h}\cdot \left(1+{\tan }^{2}x\right)={\sec }^{2}x$.

$=f^{\prime }(x)={\lim }_{h\to 0}\frac{\tan h}{h}\cdot \left(1+{\tan }^{2}x\right)={\sec }^{2}x.$

(ii) $f^{\prime }(x)={\lim }_{h\to 0}\frac{e^{\sin (x+h)}-e^{\sin x}}{h}={\lim }_{h\to 0}{e}^{\sin x}\frac{\left[e^{\sin (x+h)-\sin x}-1\right]}{\sin (x+h)-\sin x}\left(\frac{\sin (x+h)-\sin x}{h}\right)$

$=e^{\sin x}{\lim }_{h\to 0}\frac{\sin (x+h)-\sin x}{h}=e^{\sin x}\cos x$

Ex.2 Differentiate each of following functions

(i) y = x² + sin x  (ii) y = e^x tan x

(iii) $y=\frac{x}{\sin x}$ (iv) y = cos (lnx)

Sol.

(i) y = x² + sin x

$\frac{dy}{dx}=\frac{d}{dx}\left(x^2+\sin x\right)=\frac{d\left(x^2\right)}{dx}+\frac{d(\sin x)}{dx}$

$\frac{dy}{dx}=2x+\cos x$

(ii) y = e^x tan x

$\frac{dy}{dx}=\frac{d}{dx}\left(e^x\tan x\right)=\tan x\cdot \frac{d\left(e^x\right)}{dx}+e^x\frac{d}{dx}(\tan x)$

$\frac{dy}{dx}=tanx.ex+exsec2x$

(iii) $y=\frac{x}{\sin x}$

$\frac{dy}{dx}=\frac{\sin x\frac{d}{dx}(x)-x\frac{d}{dx}(\sin x)}{{\sin }^{2}x}$

$\frac{dy}{dx}=\frac{\sin x-x\cos x}{{\sin }^{2}x}$

(iv) y = cos (lnx)

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(cos(lnx))$

$=–sin(lnx).\frac{\mathrm{d}}{\mathrm{dx}}(lnx)$

$=–sin(lnx)\times \frac{1}{x}$

$=-\frac{\sin (\ln x)}{x}$

Ex.3 If $y={\log }_{\mathrm{e}}\left({\tan }^{-1}\sqrt{1+x^2}\right)$, find $\frac{\mathrm{dy}}{\mathrm{dx}}.$

Sol.

$y={\log }_e\left({\tan }^{-1}\sqrt{1+x^2}\right)$

On differentiating we get.

$=\frac{1}{{\tan }^{-1}\sqrt{1+x^2}}\cdot \frac{1}{1+{\left(\sqrt{1+x^2}\right)}^2}\cdot \frac{1}{2\sqrt{1+x^2}}\cdot 2x$

$=\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\left\{1+{\left(\sqrt{1+x^2}\right)}^2\right\}\sqrt{1+x^2}}=\frac{x}{\left({\tan }^{-1}\sqrt{1+x^2}\right)\left(2+x^2\right)\sqrt{1+x^2}}$

Ex.4 Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ if y = x^x

Sol.

y = x^x

Taking log on both side, we get 

log y = log (x^x)

log y = x log x

on differentiating, we get 

$\frac{1}{y}\frac{dy}{dx}=\log x+x\times \frac{1}{x}$

$\frac{dy}{dx}=y[\log x+1]$

$\frac{dy}{dx}=x^x[\log x+1]$

Ex.5 If $y=\frac{x^{1/2}(1-2x{)}^{2/3}}{(2-3x{)}^{3/4}(3-4x{)}^{4/5}}$ find $\frac{\mathrm{dy}}{\mathrm{dx}}$

Sol.

ln $y=\frac{1}{2}lnx+\frac{2}{3}ln(1–2x)–\frac{3}{4}ln(2–3x)–\frac{4}{5}ln(3–4x)$

On differentiating we get,

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{2x}-\frac{4}{3(1-2x)}+\frac{9}{4(2-3x)}+\frac{16}{5(3-4x)}$

$\frac{dy}{dx}=y\left(\frac{1}{2x}-\frac{4}{3(1-2x)}+\frac{9}{4(2-3x)}+\frac{16}{5(3-4x)}\right)$

Ex.6 If x^y + y^x = 2, then find $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Sol.

Let u = x^y and v = y^x

u + v = 2   ⇒ $\frac{du}{dx}+\frac{dv}{dx}=0$

Now u = x^y   and v = y^x

⇒ ln u = y lnx  and  ln v = x ln y

$\frac{1}{u}\frac{du}{dx}=\frac{y}{x}+\ln x\frac{dy}{dx}$ and  

$\frac{1}{v}\frac{dv}{dx}=\ln y+\frac{x}{y}\frac{dy}{dx}$

$\Rightarrow \frac{du}{dx}=x^y\left(\frac{y}{x}+\ln x\frac{dy}{dx}\right)and\frac{dv}{dx}=y^x\left(\ln y+\frac{x}{y}\frac{dy}{dx}\right)$

$\Rightarrow x^y\left(\frac{y}{x}+\ln x\frac{dy}{dx}\right)+y^x\left(\ln y+\frac{x}{y}\frac{dy}{dx}\right)=0$

$\Rightarrow \frac{dy}{dx}=-\frac{\left(y^x\ln y+x^y\cdot \frac{y}{x}\right)}{\left(x^y\ln x+y^x\cdot \frac{x}{y}\right)}$

Aliter :

φ(x, y) = x^y + y^x – 2 = 0

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\partial \varphi /\partial \mathrm{x}}{\partial \varphi /\partial \mathrm{y}}=-\left(\frac{{\mathrm{yx}}^{\mathrm{n}-1}+{\mathrm{y}}^{\mathrm{x}}\ln \mathrm{y}}{{\mathrm{x}}^{\mathrm{y}}\ln \mathrm{x}+{\mathrm{xy}}^{\mathrm{x}-1}}\right)$

Ex.7 Differentiation log_e (tan x) with respect to sin^–1(e^x)

Sol.

$\frac{d\left({\log }_e\tan x\right)}{d({\sin }^{-1}\left(e^x\right)}=\frac{\cot x\cdot {\sec }^{2}x}{e^x\cdot 1/\sqrt{1-e^{2x}}}=\frac{e^{-x}\sqrt{1-e^{2x}}}{\sin x\cos x}$

Ex.8 If f(x) = 2^x + x³ + 1 and g(x) = f^–1(x), then find g’(2)

Sol.

$g(x)=f–1(x)$

$g(f(x))=x$

$g’(f(x)).f’(x)=1$

$g’(f(x)=\frac{1}{f^{\prime }(x)}=\frac{1}{2^x\ln 2+3x^2}$

$f(x)=2$

$2^x+x^3+1=2\Rightarrow x=0$

$g^{\prime }(f(0))=\frac{1}{f^{\prime }(0)}=\frac{1}{\ln 2}$

$g^{\prime }(2)=\frac{1}{\ln 2}$

Ex.9 If x = a(t + sin t) and y = a(1 – cos t), find $\frac{d^{2}y}{d{x}^2}$.

Sol.

Here x = a(t + sin t) and y = a(1 – cos t)

Differentiating both sides w.r.t., t we get :

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}(1+\cos \mathrm{t})$ and$\frac{dy}{dt}=a(\sin t)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{a}\sin \mathrm{t}}{\mathrm{a}(1+\cos \mathrm{t})}=\frac{2\sin \frac{\mathrm{t}}{2}\cdot \cos \frac{\mathrm{t}}{2}}{2{\cos }^2\frac{\mathrm{t}}{2}}=\tan \left(\frac{\mathrm{t}}{2}\right)$

Again differentiating both sides, we get,

$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\sec }^2\left(\frac{\mathrm{t}}{2}\right)\cdot \frac{1}{2}\cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{2}{\sec }^2(\mathrm{t}/2)\cdot \frac{1}{\mathrm{a}(1+\cos \mathrm{t})}=\frac{1}{2\mathrm{a}}\cdot \frac{{\sec }^2\left(\frac{\mathrm{t}}{2}\right)}{2\left({\cos }^2\frac{\mathrm{t}}{2}\right)}$

Hence, $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{1}{4\mathrm{a}}\cdot {\sec }^4\left(\frac{\mathrm{t}}{2}\right)$

Ex.10 Find ${\lim }_{x\to 0}\frac{\sin x-x}{x^3}$

Sol.

${\lim }_{x\to 0}\frac{\sin x-x}{x^3}\left(\frac{0}{0}\right)$

applying L’ Hopital rule

${\lim }_{x\to 0}\frac{\cos x-1}{3x^2}={\lim }_{x\to 0}-\frac{1}{3}\frac{(1-\cos x)}{x^2}=-\frac{1}{6}$

5.0Practice Problem on Differentiation

(1) Find the derivative of y = cos(x²) using first principle.

(2) If f(x) = (1 + x)(3 + x²)^1/2, then f'(–1) is equal to-

(A) 0 (B) $2\sqrt{2}$ (C) 4 (D) 6

(3) If $y=\frac{x^{12}+x^6+1}{x^6+x^3+1}\&\frac{dy}{dx}=x^2\left(a{x}^3+b\right)$, then the value of (a + b) is equal to

(4) If f(x) = (x + 1)(x + 2) ……. (x + n) the f'(0) is

(5) If x = secθ & y = secⁿθ – cosⁿθ, then show that $\left(x^2+4\right){\left(\frac{dy}{dx}\right)}^2=n^2\left(y^2+4\right)$

(6) If $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\dots \dots \infty }}}$, then $\frac{dy}{dx}(\sin x>0)$ , is equal to-

(A) $\frac{\cos x}{2y-1}$ (B) $\frac{\cos x}{\sqrt{1+4\sin x}}$

(C) $\frac{y\cos x}{y^2+\sin x}$ (D) $\frac{y\cos x}{y^2-\sin x}$

(7) If x = 2cos t – cos 2t ; y = 2sin t – sin 2t, find $\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{dx}{}^2}$ at $\mathrm{t}=\frac{\pi }{2}.$

(8) $\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\sin }^2\left({\cot }^{-1}\sqrt{\frac{1+\mathrm{x}}{1-\mathrm{x}}}\right)\right\}=$

(A) $-\frac{1}{2}$ (B) 0 (C) $\frac{1}{2}$ (D) –1

(9) If f(x) = x + 3x³ + 5x⁵ and g = f^–1, then find g'(9) and g"(9)

(10) Prove that ${\lim }_{x\to 0^{+}}{x}^x=1$

Answers:

(1) –2x sin(x²)

(2) C  

(3) 3

(4) $\mathrm{n}!\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\dots ..+\frac{1}{\mathrm{n}}\right)$

(6) A, B, C

(7) $-\frac{3}{2}$

(8) A

(9) $g^{\prime }(9)=\frac{1}{35},g^{\prime \prime }(9)=\frac{118}{(35{)}^3}$

6.0Sample Questions on Differentiation

Q.1 What is differentiation?

Ans. The instantaneous rate of change of a function with respect to another quantity is called differentiation.

If y = f(x) is a differentiable function of x, then

$\frac{dy}{dx}=f^{\prime }(x)={\lim }_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

Q.2 What are the differentiation Rules in calculus?

Ans. If f and g are derivable functions of x, then

(a) $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}\pm \mathrm{g})=\frac{\mathrm{df}}{\mathrm{dx}}\pm \frac{\mathrm{dg}}{\mathrm{dx}}$

(b) $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{cf})=\mathrm{c}\frac{\mathrm{df}}{\mathrm{dx}}$, where c is any constant

(c) $\frac{d}{dx}(fg)=f\frac{dg}{dx}+g\frac{df}{dx}$ where g ≠ 0 known as “PRODUCT RULE”

(d) $\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(\frac{\mathrm{df}}{\mathrm{dx}}\right)-\mathrm{f}\left(\frac{\mathrm{dg}}{\mathrm{dx}}\right)}{{\mathrm{g}}^2}$ where g ≠ 0 known as “QUOTIENT RULE”

(e) If y = f(u) & u = g(x) then $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ known as “CHAIN RULE”