Implicit Differentiation

In this section, ask-math will explain you about Implicit differentiation
A function can be implicit or explicit.

Explicit form
When 'y' is expressed as function of x then such a function is called Explicit function.
For example : 1) y = $x^{2}$ +1
2) y = sin(x) + $x^{3}$
3) y = $\sqrt{x+1}$ 		Implicit form
When the function is expressed in terms of x and y then such equations are called Implicit form.
For example : 1) $x^{2} + y^{2}$= 5
2) $x^{2}$ + xy = 0
3) $y^{3} -y^{2} + x^{2}$= 0

Implicit differentiation : So far we have discussed derivatives of the function y = f(x). If the derivatives x and y are connected by a relation of the form f(x,y)= 0 and sometimes it is not possible to express y as a function of x in the form of y = g(x) then y is said to an implicit function of x. To find dy/dx in such a case, we differentiate both sides of the given relation with respect to x. So to differentiate such functions we use a chain rule.

Examples on Implicit differentiation

Example 1 : If $x^{2} + 2xy +y^{3}$= 42, find $\frac{dy}{dx}$

Solution : We have $x^{2} + 2xy +y^{3}$= 42
Differentiating both sides with respect to x, we get

$\frac{d}{dx}(x^{2})+ 2 \frac{d}{dx}(xy) + \frac{d}{dx}(y^{3}) = \frac{d}{dx}(42)$

Apply the power rule, product rule and chain rule for differentiation
2x + 2(x.$\frac{dy}{dx} + y.1) + 3y^{2}\frac{dy}{dx}$ = 0

2x + 2x. $\frac{dy}{dx} + 2y + 3y^{2}\frac{dy}{dx}$ = 0

2x + 2y + $\frac{dy}{dx}(2x + 3y^{2})$ = 0

$\frac{dy}{dx}(2x + 3y^{2})$ = -2x - 2y

$\frac{dy}{dx}(2x + 3y^{2})$ = -(2x + 2y)

$\frac{dy}{dx} = \frac{-(2x + 2y)}{(2x + 3y^{2})}$

Example 2 : If sin(y) = x.sin(a + y) , find $\frac{dy}{dx}$

Solution : We have sin(y) = x.sin(a + y)
Differentiating both sides with respect to x, we get
$\frac{d}{dx}(sin(y))= \frac{d}{dx}[x.sin(a + y)] $

Apply the product rule and chain rule for differentiation
cos(y)$\frac{dy}{dx}= 1.sin(a + y) + x.cos(a+ y) \frac{d}{dx}(a + y)$

cos(y)$\frac{dy}{dx}= sin(a + y) + x.cos(a+ y)[ \frac{d}{dx}(a) + \frac{dy}{dx}] $

cos(y)$\frac{dy}{dx}= sin(a + y) + x.cos(a+ y) . \frac{dy}{dx} $

cos(y)$\frac{dy}{dx} - x.cos(a+ y) . \frac{dy}{dx} $ = sin(a + y)

$\frac{dy}{dx}[cos(y) - x.cos(a + y)]$ = sin(a + y)

$\frac{dy}{dx} = \frac{sin(a + y)}{[cos(y) - x.cos(a + y)]}$

12th grade math

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