Implicit Differentiation, Example’s Solution

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Implicit Differentiation:

Implicit functions: If $y$ and $x$ are mixed up and $y$ cannot be expressed in terms of the independent variable $x$ , Then $y$ is called an Implicit functions. Symbolically it is written as $f\left(x,y\right)=0$

Differentiation of implicit function

Examples:

$x^3+xy-x=7$

$y^3+xy-x=0$

$y^3+xy-xy^3=0$

Procedure:

Step (1) when $x$ and $y$ are related Implicit we assumethat $y$ is differentiable function of $x$
Step (2) Differentiate both sides of eq w.r.t $x$
Step (3) Solve the resulting eq for $\frac{dy}{dx}$

Example 1: Implicit Differentiation

$x^3+xy-x=7$

Differentiate w.r.t $x$

$\frac d{dx}\left(x^3+xy-x\right)=\frac d{dx}7$

$\frac d{dx}\left(x^3\right)+\frac d{dx}\left(xy\right)-\frac d{dx}(x)=\frac d{dx}7$ [sum and difference rule]

$3x^2+\left[y\frac d{dx}x+x\frac d{dx}y\right]-1=0$

$3x^2+y+x\frac d{dx}y=1$

$x\frac d{dx}y=1-3x^2-y$

$\boxed{\frac d{dx}y=\frac{1-3x^2-y}x}$

Example 2: [implicit Differentiation]

$y^3+xy-x=0$

Differentiate w.r.t $x$

$\frac d{dx}\left(y^3+xy-x\right)=\frac d{dx}0$

$\frac d{dx}y^3+\frac d{dx}xy-\frac d{dx}x=\frac d{dx}0$ [sum and difference rule]

$3y^2+y\frac d{dx}x+x\frac d{dx}y-1=0$

$3y^2+y+x\frac d{dx}y=1$

$x\frac d{dx}y=1-3y^2-y$

$\boxed{\frac d{dx}y=\frac{1-3y^2-y}x}$

Example 3: [implicit Differentiation]

$y^3+y-xy^3=0$

Differentiate w.r.t $x$

$\frac d{dx}\left(y^3+y-xy^3\right)=\frac d{dx}0$

$\frac d{dx}y^3+\frac d{dx}y-\frac d{dx}xy^3=0$ [sum and difference rule]

$3y^2+\frac d{dx}y-\left[y^3\frac d{dx}x+x\frac d{dx}y^3\right]=0$

$3y^2+\frac d{dx}y-\left[y^3+3xy^2\frac{dy}{dx}\right]=0$

$3y^2+\frac d{dx}y-y^3-3xy^2\frac{dy}{dx}=0$

$\frac d{dx}y-3xy^2\frac{dy}{dx}=y^3-3y^2$

$\frac d{dx}\left(y-3xy^2\right)=y^3-3y^2$

$\frac{dy}{dx}=\frac{y^3-3y^2}{y-3xy^2}$

$\frac{dy}{dx}=\frac{y\left(y^2-3y\right)}{y(1-3xy)}$

$\boxed{\frac{dy}{dx}=\frac{y^2-3y}{1-3xy}}$

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