Chapter 3.6, Problem 1E

To determine

To calculate $y^{\prime }$ , in terms of x , where

  $y=u^2;\text{ and }u=x^3-4$

Answer to Problem 1E

  $y^{\prime }=6x^2\left(x^3-4\right)$

Explanation of Solution

Given:

  $y=u^2;\text{ and }u=x^3-4$

Concept Used:

• Chain rule of differentiation is: $\frac{d}{dx}f\left[g\left(x\right)\right]=\frac{d}{dg}f\left(g\right)\frac{dg}{dx}$ , where g is a differentiable function at x and f is a differentiable function at $g\left(x\right)$
• Power rule of differentiation formula

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Calculation:

To calculate $y^{\prime }$ , in terms of x , where

  $y=u^2;\text{ and }u=x^3-4$

First, put value of u in y as

  $y={\left(x^3-4\right)}^2$

Now, using chain rule of differentiation differentiate above, as

  $\begin{array}{l}\frac{d}{dx}y=\frac{d}{dx}{\left(x^3-4\right)}^2\\ y^{\prime }=2\left(x^3-4\right)\frac{d}{dx}\left(x^3-4\right)\\ y^{\prime }=2\left(x^3-4\right)\left(\frac{d}{dx}{x}^3-\frac{d}{dx}4\right)\\ y^{\prime }=2\left(x^3-4\right)\left(3x^2-0\right)\\ \Rightarrow y^{\prime }=6x^2\left(x^3-4\right)\end{array}$

Hence,

  $y^{\prime }=6x^2\left(x^3-4\right)$