Rules for differentiation

Suppose , c is a constant and u , v are functions of x .

1. $ \displaystyle \frac{d}{dx}(c) = 0 $

2. $ \displaystyle \frac{d}{dx}(c u) = c\frac{du}{dx} $

3. $ \displaystyle \frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx} $

4. $ \displaystyle \frac{d}{dx}(u – v) = \frac{du}{dx} – \frac{dv}{dx} $

5. $ \displaystyle \frac{d}{dx}(u . v) = u \frac{dv}{dx} + v \frac{du}{dx} $

6. $ \displaystyle \frac{d}{dx}(\frac{u}{v}) = \frac{v \frac{du}{dx} – u \frac{dv}{dx}}{v^2} $

Illustration : Find dy/dx If

(i) $ \displaystyle y = (a\sqrt{x} + b) (b\sqrt{x}+a) $

(ii) $ \displaystyle y = \frac{1 – cosx}{1 + cosx} $

Solution:

(i) $ \displaystyle y = (a\sqrt{x} + b) (b\sqrt{x}+a) $

$ \displaystyle \frac{dy}{dx} = (a\sqrt{x} + b) \frac{d}{dx}(b\sqrt{x}+a) + (b\sqrt{x}+a)\frac{d}{dx}(a\sqrt{x} +b) $

$ \displaystyle = (a\sqrt{x} + b) b\frac{1}{2\sqrt{x}} + (b\sqrt{x} + a)a \frac{1}{2\sqrt{x}} $

$ \displaystyle = ab + \frac{a^2 + b^2}{2\sqrt{x}} $

(ii) $ \displaystyle y = \frac{1 – cosx}{1 + cosx} $

$ \displaystyle \frac{dy}{dx} = \frac{d}{dx} (\frac{1-cosx}{1+cosx}) $

$ \displaystyle = \frac{(1+cosx)\frac{d}{dx}(1-cosx)-(1-cosx)\frac{d}{dx}(1+cosx)}{(1+cosx)^2} $

$ \displaystyle = \frac{(1+cosx)(sinx)-(1-cosx)(-sinx)}{(1+cosx)^2} $

$ \displaystyle = \frac{sinx(1+cosx)+ sinx(1-cosx)}{(1+cosx)^2} $

$ \displaystyle = \frac{2sinx}{(1+cosx)^2} $

Illustration : (i) If y logx = x − y , prove that

$ \displaystyle \frac{dy}{dx} = \frac{logx}{(1+logx)^2} $

(ii) If siny = x sin(a + y) , prove that

$ \displaystyle \frac{dy}{dx} = \frac{sina}{x^2 – 2 x cosa +1 } $

Solution:

(i) y logx = x − y

$ \displaystyle y(1+logx)= x $

$ \displaystyle y= \frac{x}{1+logx} $

$ \displaystyle \frac{dy}{dx} = \frac{(1+logx)\frac{d}{dx}(x) – x\frac{d}{dx}(1+logx)}{(1+logx)^2} $

$ \displaystyle = \frac{(1+logx)-x.\frac{1}{x}}{(1+logx)^2} $

$ \displaystyle = \frac{logx}{(1+logx)^2} $

(ii) We have siny = x sin(a + y)

siny = x(sina.cosy + cosa.siny)

xsina.cosy + (xcosa − 1) siny = 0 …. (1)

Differentiate w.r.t.x we get

$\large [sina + (xcosa-1)\frac{dy}{dx}]cosy + (cosa-xsina \frac{dy}{dx})sina$ ……(2)

By eliminating siny and cosy from (1) and (2) we get

$ \large  \left| \begin{array}{cc} x sina & x cosa-1  \\ sina + (xcosa -1)\frac{dy}{dx} & cosa – x sina\frac{dy}{dx} \end{array} \right| = 0 $

$\large \frac{dy}{dx} = \frac{sina}{x^2 – 2xcosa + 1}$

Exercise :

Find the derivative w.r.t. x. of the following functions

(i) $\large (1+3x^2) (1+5x)$

(ii) $\large \frac{x+1}{\sqrt{x+1}}$

iii) $\large sin(2x-1) cos(x+1)$

(iv) $\large \frac{tanx-1}{secx}$

(v) $\large (\frac{x}{1+tanx})^{1/2}$

(vi) $\large \sqrt{\frac{secx+tanx}{secx-tanx}}$

Differentiation of function of function

If y = f(u) and u = g(x), then $\large \frac{dy}{dx}= \frac{dy}{du}.\frac{du}{dx}$

Illustration : Find $\large \frac{dy}{dx}$

(i) $\large sin\sqrt{cosx}$

(ii) $\large sin(log\sqrt{\frac{x}{x+1}})$

(iii) $\large log(x+ e^{\sqrt{x}})$

Solution: (i)

$\large \frac{dy}{dx}= \frac{d(sin\sqrt{cosx})}{\sqrt{cosx}} .\frac{d(\sqrt{cosx})}{d(cosx)} . \frac{d(cosx)}{dx}$

$\large = cos\sqrt{cosx} . \frac{1}{2\sqrt{cosx}} .(-sinx)$

$\large = -\frac{sinx .cos\sqrt{cosx}}{2\sqrt{cosx}}$

Differentiation of implicit functions:

Given the equation f(x,y) = c

For finding dy/dx , we differentiate both the sides w.r.t, x, treating y as a function of x , and then solve equation for dy/dx.

Illustration : Find dy/dx

(i) log(xy) = x2 + y2

(ii) x+y = sin(xy)

Solution: (i) log(xy) = x2 + y2

=> logx + logy = x2 + y2

Differentiating w. r. to x ,

$\large \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$

$\large (\frac{1}{y}-2y)\frac{dy}{dx} = 2x-\frac{1}{x}$

$\large \frac{1-2y^2}{y}\frac{dy}{dx} = \frac{2x^2 – 1}{x}$

$\large \frac{dy}{dx} = \frac{y(2x^2-1)}{x(1-2y^2)}$

(ii) x + y = sin(xy)

Differentiating w.r.t. x, we get

$\large 1 + \frac{dy}{dx} = cos(xy)(x\frac{dy}{dx} + y)$

$\large \frac{dy}{dx} (1-xcos(xy)) = y cos(xy) – 1$

$\large \frac{dy}{dx} = \frac{y cos(xy)-1}{1-xcos(xy)}$

Differentiation of parametric functions :

Working Rule:

(i) If x and y are function of parameter t , then find dx/dt and dy/dt separately.

(ii) Then $\large \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $

e.g., x = a(θ + sinθ) , y = a(1 − cosθ) where θ is parameter.

$\large \frac{dx}{d\theta}= a(1+cos\theta) $

$\large \frac{dy}{d\theta}= a(0 + sin\theta)= a sin\theta $

$\large \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} $

$\large \frac{dy}{dx} = \frac{a sin\theta}{a(1+cos\theta)}$

$\large \frac{dy}{dx} = \frac{2sin(\theta/2) cos(\theta/2)}{2cos^2(\theta/2)}$

$\large \frac{dy}{dx} = tan(\theta/2)$

Also Read

→Limits of a Function
→Continuity of a Function
→Differentiability of a Function
→Rules for differentiation
→Higher order derivatives
→L’HOSPITAL’S RULE

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