Chapter 3.5, Problem 10E

To determine

To calculate: The derivative of the function $1+x=\sin \left(x{y}^2\right)$ by implicit differentiation.

Answer to Problem 10E

The derivative of the function is $\frac{1-y^2\cos \left(x{y}^2\right)}{2xy\cos \left(x{y}^2\right)}$ .

Explanation of Solution

Given information:

The function $1+x=\sin \left(x{y}^2\right)$ .

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 $1+x=\sin \left(x{y}^2\right)$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(1+x\right)=\frac{d}{dx}\left(\sin \left(x{y}^2\right)\right)\\ \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(\sin \left(x{y}^2\right)\right)\end{array}$

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,

  $\begin{array}{c}\frac{d}{dx}\left(1+x\right)=\frac{d}{dx}\left(\sin \left(x{y}^2\right)\right)\\ \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left(\sin \left(x{y}^2\right)\right)\\ 0+1=\cos \left(x{y}^2\right)\left[y^2+2xyy\text{'}\right]\\ 1=y^2\cos \left(x{y}^2\right)+2xy\cos \left(x{y}^2\right)y\text{'}\end{array}$

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

  $\begin{array}{c}{y}^2\cos \left(x{y}^2\right)+2xy\cos \left(x{y}^2\right)y\text{'}=1\\ 2xy\cos \left(x{y}^2\right)y\text{'}=1-y^2\cos \left(x{y}^2\right)\\ y\text{'}=\frac{1-y^2\cos \left(x{y}^2\right)}{2xy\cos \left(x{y}^2\right)}\end{array}$

Thus, the derivative of the function is $\frac{1-y^2\cos \left(x{y}^2\right)}{2xy\cos \left(x{y}^2\right)}$ .