Chapter 3.5, Problem 7E

To determine

To calculate: The derivative of the function $x^4\left(x+y\right)=y^2\left(3x-y\right)$ by implicit differentiation.

Answer to Problem 7E

The derivative of the function is $\frac{3y^2-5x^4-4x^{3}y}{x^4-6xy+3y^2}$ .

Explanation of Solution

Given information:

The function $x^4\left(x+y\right)=y^2\left(3x-y\right)$ .

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 $x^4\left(x+y\right)=y^2\left(3x-y\right)$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left(x^4\left(x+y\right)\right)=\frac{d}{dx}\left(y^2\left(3x-y\right)\right)\\ \frac{d}{dx}\left(x^5\right)+\frac{d}{dx}\left(x^{4}y\right)=\frac{d}{dx}\left(3x{y}^2\right)-\frac{d}{dx}\left(y^3\right)\end{array}$

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,

  $\begin{array}{c}\frac{d}{dx}\left(x^4\left(x+y\right)\right)=\frac{d}{dx}\left(y^2\left(3x-y\right)\right)\\ \frac{d}{dx}\left(x^5\right)+\frac{d}{dx}\left(x^{4}y\right)=\frac{d}{dx}\left(3x{y}^2\right)-\frac{d}{dx}\left(y^3\right)\\ 5x^4+x^{4}y\text{'}+4x^{3}y=3y^2+6xyy\text{'}-3y^{2}y\text{'}\end{array}$

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

  $\begin{array}{c}5x^4+x^{4}y\text{'}+4x^{3}y=3y^2+6xyy\text{'}-3y^{2}y\text{'}\\ \left(x^4-6xy+3y^2\right)y\text{'}=3y^2-5x^4-4x^{3}y\\ y\text{'}=\frac{3y^2-5x^4-4x^{3}y}{x^4-6xy+3y^2}\end{array}$

Thus, the derivative of the function is $\frac{3y^2-5x^4-4x^{3}y}{x^4-6xy+3y^2}$ .