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Implicit differentiation – Advanced Examples

  • Last Updated : 28 Jan, 2021

In the previous article, we have discussed the introduction part and some basic examples of Implicit differentiation. So in this article, we will discuss some advanced examples of implicit differentiation.

Implicit Differentiation

Implicit differentiation is a method that makes use of the chain rule to differentiate implicitly defined functions. It is generally not easy to find the function explicitly and then differentiate. Instead, we can totally differentiate f(x, y) and then solve the rest of the equation to find the value of \frac{dy}{dx}. Even when it is possible to explicitly solve the original equation, the formula resulting from total differentiation is, in general, much simpler and easier to use.

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Method to solve

  • Differentiate both sides of the equation with respect to x.
  • Follow the rules of differentiation.
  • Use the chain rule to differentiate expressions involving y.
  • Solve the equation for \frac{dy}{dx}

Advanced Examples 

Example 1: Find the derivative of y = cos(5x – 3y)?



Solution:

Given equation:

                      y = cos(5x – 3y)

Step 1: Differentiating both sides wrt x,

                     \frac{d}{dx}y = \frac{d}{dx}(cos(5x - 3y))

Step 2: Using Chain Rule

                    \frac{dy}{dx} = -sin(5x-3y)\frac{d}{dx}(5x - 3y) \\ \\ \qquad \qquad \ \ \frac{dy}{dx} = -sin(5x-3y)(5 - 3\frac{dy}{dx})

Step 3: Expanding the above equation



                     \frac{dy}{dx} = -5sin(5x-3y)+ 3sin(5x-3y)\frac{dy}{dx}

Step 4: Taking all terms with dy/dx on LHS

                    \frac{dy}{dx} - 3sin(5x-3y)\frac{dy}{dx}= -5sin(5x-3y)

Step 5: Taking dy/dx common from the LHS of equation

                    \frac{dy}{dx}(1 - 3sin(5x-3y))= -5sin(5x-3y)

Step 6: Isolate dy/dx

                    \frac{dy}{dx}= \frac{-5sin(5x-3y)}{1 - 3sin(5x-3y)}

Example 2: Find the derivative of (x² + y²)³ = 5x²y²?

Solution:

Given equation:



(x² + y²)³ = 5x²y²

Differentiating both sides:

\begin{aligned} \frac{d}{dx}(x^2+y^2)^3 &=\frac{d}{dx}5x^2y^2 \\ 3(x^2+y^2)^2\frac{d}{dx}(x^2+y^2) &=5\frac{d}{dx}x^2y^2 \\ 3(x^2+y^2)^2(2x+2y\frac{dy}{dx}) &=5(2xy^2+2y\frac{dy}{dx}x^2) \\ 6x(x^2+y^2)^2+6y(x^2+y^2)^2\frac{dy}{dx} &=10xy^2+10x^2y\frac{dy}{dx} \\ 6y(x^2+y^2)^2\frac{dy}{dx}-10x^2y\frac{dy}{dx} &=10xy^2-6x(x^2+y^2)^2 \\ (6y(x^2+y^2)^2-10x^2y)\frac{dy}{dx} &=10xy^2-6x(x^2+y^2)^2 \\ \frac{dy}{dx} &=\frac{10xy^2-6x(x^2+y^2)^2}{6y(x^2+y^2)^2-10x^2y} \\ \end{aligned}

Example 3: Find the derivative of  e^{xy²} = x-y?

Solution:

Given equation:

e^{xy²} = x-y

Differentiating both sides:

\begin{aligned} \frac{d}{dx}e^{xy²}&=\frac{d}{dx}(x-y) \\ e^{xy²}\frac{d}{dx}(xy²)&=1-\frac{dy}{dx} \\ e^{xy²}(y^2+x.2y\frac{dy}{dx})&=1-\frac{dy}{dx} \\ y^2e^{xy²}+2xye^{xy²}\frac{dy}{dx}&=1-\frac{dy}{dx} \\ e^{xy²}2xy\frac{dy}{dx}+\frac{dy}{dx}&=1-y^2e^{xy²} \\ (2xye^{xy²}+1)\frac{dy}{dx}&=1-y^2e^{xy²} \\ \frac{dy}{dx}&=\frac{1-y^2e^{xy²}}{2xye^{xy²}+1} \\ \frac{dy}{dx}&=\frac{1-y^2(x-y)}{2xy(x-y)+1} \\ \frac{dy}{dx}&=\frac{1-xy^2+y^3}{2x^2y-2xy^2+1} \\ \end{aligned}

Example 4: Find the derivative of y = ln(x)?

Solution:

Given equation:

y = ln(x)

=> ey = x

Differentiating both sides:

\begin{aligned} \frac{d}{dx}e^{y}&=\frac{d}{dx}(x)  \\ e^{y}\frac{dy}{dx}&=1 \\ \frac{dy}{dx}&=\frac{1}{e^y} \\ \frac{dy}{dx}&=\frac{1}{x} \\ \end{aligned}




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