# JEE Main & Advanced Mathematics Differentiation Methods of Differentiation

## Methods of Differentiation

Category : JEE Main & Advanced

(1) Differentiation of implicit functions : If $y$ is expressed entirely in terms of $x,$ then we say that $y$ is an explicit function of $x$. For example $y=\sin \,\,x,y=\text{ }{{e}^{x}},\,\,\,y={{x}^{2}}+x+1$ etc. If $y$ is related to $x$ but cannot be conveniently expressed in the form of $y=f(x)$ but can be expressed in the form $f(x,y)=0$, then we say that $y$ is an implicit function of $x$.

Working Rule 1:

(a) Differentiate each term of $f(x,y)=0$ with respect to $x$.

(b) Collect the terms containing $dy/dx$ on one side and the terms not involving $dy/dx$ on the other side.

(c) Express $dy/dx$ as a function of $x$ or $y$ or both.

In case of implicit differentiation, $dy/dx$ may contain both $x$ and $y$.

Working Rule 2:

If $f(x,\,\,y)=$ constant, then $\frac{dy}{dx}=-\frac{\left( \frac{\partial f}{\partial x} \right)}{\left( \frac{\partial f}{\partial y} \right)}$, where $f(x)$ and $\frac{\partial f}{\partial y}$ are partial differential coefficients of $f(x,\,y)$ with respect to $x$ and $y$ respectively.

(2) Logarithmic differentiation : If differentiation of an expression or an equation is done after taking log on both sides, then it is called logarithmic differentiation. This method is useful for the function having following forms

(i) $y={{[f(x)]}^{g(x)}}$

(ii) $y=\frac{{{f}_{1}}(x).{{f}_{2}}(x).........}{{{g}_{1}}(x).{{g}_{2}}(x)........}$ where ${{g}_{i}}(x)\ne 0\,$

(where $i=\text{ }1,\text{ }2,\text{ }3,.....$), ${{f}_{i}}(x)$ and ${{g}_{i}}(x)$ both are differentiable.

(i) Case I : $y=[f{{(x]}^{g(x)}}$ where $f(x)$ and $g(x)$are functions of $x$. To find the derivative of this type of functions we proceed as follows: Let $y={{[f(x)]}^{g(x)}}$. Taking logarithm of both the sides, we have $]\,a,\,b[,$$f(x)$and then we differentiate w.r.t. $x$.

(ii) Case II : $y=\frac{{{f}_{1}}(x).{{f}_{2}}(x)}{{{g}_{1}}(x).{{g}_{2}}(x)}$

Taking logarithm of both the sides, we have

$\log y=\log [{{f}_{1}}(x)]+\log [{{f}_{2}}(x)]-\log [{{g}_{1}}(x)]-\log [{{g}_{2}}(x)]$

and differentiating w.r.t. $x,$ we get

$\frac{1}{y}\frac{dy}{dx}=\frac{{{{{f}'}}_{1}}(x)}{{{f}_{1}}(x)}+\frac{{{{{f}'}}_{2}}(x)}{{{f}_{2}}(x)}-\frac{{{{{g}'}}_{1}}(x)}{{{g}_{1}}(x)}-\frac{{{{{g}'}}_{2}}(x)}{{{g}_{2}}(x)}$

(3) Differentiation of parametric functions : Sometimes $x$ and $y$ are given as functions of a single variable, e.g., $x=\phi (t),\,\,y=\psi (t)$ are two functions and t is a variable. In such a case $x$ and $y$ are called parametric functions or parametric equations and t is called the parameter. To find $\frac{dy}{dx}$ in case of parametric functions, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.

(4) Differentiation of infinite series : If $y$ is given in the form of infinite series of $x$ and we have to find out $]\,a,\,b[,$, then we remove one or more terms, it does not affect the series.

(i) If $y=\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+....\infty }}}$, then $y=\sqrt{f(x)+y}$

$\Rightarrow$${{y}^{2}}=f(x)+y$$\Rightarrow$ $2y\frac{dy}{dx}={f}'(x)+\frac{dy}{dx}$; $\therefore$$\frac{dy}{dx}=\frac{{f}'(x)}{2y-1}$.

(ii) If $y=f{{(x)}^{f(x)}}^{f{{(x)}^{f(x).....\infty }}}$ then $y=f{{(x)}^{y}}$

$\therefore$ $\log y=y\log f(x)$;  $\frac{1}{y}\frac{dy}{dx}=\frac{y.{f}'(x)}{f(x)}+\log f(x).\frac{dy}{dx}$

$\therefore$ $\frac{dy}{dx}=\frac{{{y}^{2}}{f}'(x)}{f(x)[1-y\log f(x)]}$

(iii) If $gof(x)$ then $\frac{dy}{dx}=\frac{y{f}'(x)}{2y-f(x)}$.

(5) Differentiation of composite function : Suppose a function is given in form of $fog(x)$ or $f[g(x)]$.

Working Rule

Differentiate applying chain rule, $\frac{d}{dx}f[g(x)]=f'[g(x)].g'(x)$

You need to login to perform this action.
You will be redirected in 3 sec