Chapter 3.5, Problem 9E

To determine

To calculate: The derivative of the function $x^2{y}^2+x\sin y=4$ by implicit differentiation.

Answer to Problem 9E

The derivative of the function is $\frac{-2x{y}^2-\sin y}{2x^{2}y+x\cos y}$ .

Explanation of Solution

Given information:

The function $x^2{y}^2+x\sin y=4$ .

Formula used:

Thechain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Calculation:

Consider the function $x^2{y}^2+x\sin y=4$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(x^2{y}^2+x\sin y\right)=\frac{d}{dx}\left(4\right)\\ \frac{d}{dx}\left(x^2{y}^2\right)+\frac{d}{dx}\left(x\sin y\right)=0\end{array}$

Recall that power rule for differentiation is $\frac{d}{dx}{x}^n=n{x}^{n-1}$ and chain rule for differentiation is if f is a function of gthen $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)=f\text{'}\left(g\left(x\right)\right)g\text{'}\left(x\right)$ .

Also for the terms of the above expression, apply the product rule for differentiation.

Recall that product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Apply it. Also observe that y is a function of x,

  $\begin{array}{c}\frac{d}{dx}\left(x^2{y}^2+x\sin y\right)=\frac{d}{dx}\left(4\right)\\ \frac{d}{dx}\left(x^2{y}^2\right)+\frac{d}{dx}\left(x\sin y\right)=0\\ 2x{y}^2+2y{x}^{2}y\text{'}+\sin y+x\cos yy\text{'}=0\end{array}$

Isolate the value of $y\text{'}$ on left hand side and simplify,

  $\begin{array}{c}2x{y}^2+2y{x}^{2}y\text{'}+\sin y+x\cos yy\text{'}=0\\ \left(2x^{2}y+x\cos y\right)y\text{'}=-2x{y}^2-\sin y\\ y\text{'}=\frac{-2x{y}^2-\sin y}{2x^{2}y+x\cos y}\end{array}$

Thus, the derivative of the function is $\frac{-2x{y}^2-\sin y}{2x^{2}y+x\cos y}$ .