Chapter 3.5, Problem 8E

To determine

To calculate: The derivative of the function $y^5+x^2{y}^3=1+y{e}^{x^2}$ by implicit differentiation.

Answer to Problem 8E

The derivative of the function is $\frac{2xy\left(e^{x^2}-y^2\right)}{5y^4+3x^2{y}^2-e^{x^2}}$ .

Explanation of Solution

Given information:

The function $y^5+x^2{y}^3=1+y{e}^{x^2}$ .

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 $y^5+x^2{y}^3=1+y{e}^{x^2}$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(y^5+x^2{y}^3\right)=\frac{d}{dx}\left(1+y{e}^{x^2}\right)\\ \frac{d}{dx}\left(y^5\right)+\frac{d}{dx}\left(x^2{y}^3\right)=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(y{e}^{x^2}\right)\end{array}$

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,

  $\begin{array}{c}\frac{d}{dx}\left(y^5+x^2{y}^3\right)=\frac{d}{dx}\left(1+y{e}^{x^2}\right)\\ \frac{d}{dx}\left(y^5\right)+\frac{d}{dx}\left(x^2{y}^3\right)=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(y{e}^{x^2}\right)\\ 5y^{4}y\text{'}+2x{y}^3+3x^2{y}^{2}y\text{'}=0+e^{x^2}y\text{'}+2xy{e}^{x^2}\end{array}$

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

  $\begin{array}{c}5y^{4}y\text{'}+2x{y}^3+3x^2{y}^{2}y\text{'}=0+e^{x^2}y\text{'}+2xy{e}^{x^2}\\ \left(5y^4+3x^2{y}^2-e^{x^2}\right)y\text{'}=-2x{y}^3+2xy{e}^{x^2}\\ y\text{'}=\frac{-2x{y}^3+2xy{e}^{x^2}}{5y^4+3x^2{y}^2-e^{x^2}}\\ y\text{'}=\frac{2xy\left(e^{x^2}-y^2\right)}{5y^4+3x^2{y}^2-e^{x^2}}\end{array}$

Thus, the derivative of the function is $\frac{2xy\left(e^{x^2}-y^2\right)}{5y^4+3x^2{y}^2-e^{x^2}}$ .