Implicit differentiation - Advanced Examples

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 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.

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}

Implicit differentiation Formula

Implicit differentiation involves differentiating an implicit equation with respect to the desired variable x, while regarding the other variables as unspecified functions dependent on x.

To differentiate an implicit function, there are two common methods:

In the first method, the implicit equation is solved for ?,, expressing it explicitly in terms of x and then differentiating y with respect to x This approach is practical when y can be easily expressed in terms of x.

In the second method, y is considered as a function of x, and both sides of the implicit equation are differentiated with respect to x.

Solved Example

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^{xy^2}&=\frac{d}{dx}(x-y) \\ e^{xy^2}\frac{d}{dx}(xy^2)&=1-\frac{dy}{dx} \\ e^{xy^2}(y^2+x \cdot 2y\frac{dy}{dx})&=1-\frac{dy}{dx} \\ y^2e^{xy^2}+2xye^{xy^2}\frac{dy}{dx}&=1-\frac{dy}{dx} \\ e^{xy^2}2xy\frac{dy}{dx}+\frac{dy}{dx}&=1-y^2e^{xy^2} \\ (2xye^{xy^2}+1)\frac{dy}{dx}&=1-y^2e^{xy^2} \\ \frac{dy}{dx}&=\frac{1-y^2e^{xy^2}}{2xye^{xy^2}+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)

=> e^y = 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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Conclusion of Implicit Differentiation

Implicit differentiation is a valuable technique used to differentiate implicitly defined functions with respect to a desired variable, typically x. It offers flexibility when explicit expressions for variables are difficult to obtain or when dealing with complex equations involving multiple variables. By treating one variable as a function of another and applying the chain rule, implicit differentiation enables the determination of derivatives even in cases where explicit differentiation is impractical.

This method is indispensable in various fields of mathematics, science, and engineering, providing a powerful tool for analyzing relationships between variables in implicit equations.