Chapter 3.5, Problem 12E

To determine

To calculate: The derivative of the function $y\sin \left(x^2\right)=x\sin \left(y^2\right)$ by implicit differentiation.

Answer to Problem 12E

The derivative of the function is $\frac{\sin \left(y^2\right)-2xy\cos \left(x^2\right)}{\sin \left(x^2\right)-2xy\cos \left(y^2\right)}$ .

Explanation of Solution

Given information:

The function $y\sin \left(x^2\right)=x\sin \left(y^2\right)$ .

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 $y\sin \left(x^2\right)=x\sin \left(y^2\right)$ .

Differentiate both sides with respect to x ,

  $\frac{d}{dx}\left(y\sin \left(x^2\right)\right)=\frac{d}{dx}\left(x\sin \left(y^2\right)\right)$

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,

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

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

  $\begin{array}{c}\left[\sin \left(x^2\right)-x\cos \left(y^2\right)2y\right]y\text{'}=\sin \left(y^2\right)-y\cos \left(x^2\right)2x\\ y\text{'}=\frac{\sin \left(y^2\right)-y\cos \left(x^2\right)2x}{\sin \left(x^2\right)-x\cos \left(y^2\right)2y}\\ y\text{'}=\frac{\sin \left(y^2\right)-2xy\cos \left(x^2\right)}{\sin \left(x^2\right)-2xy\cos \left(y^2\right)}\end{array}$

Thus, the derivative of the function is $\frac{\sin \left(y^2\right)-2xy\cos \left(x^2\right)}{\sin \left(x^2\right)-2xy\cos \left(y^2\right)}$ .