Q. 41 - Calculus

Question

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating. 

$f\left(x\right)={\left(5x^4-3x^2\right)}^7\left(2x^3+1\right)$

Step-by-Step Solution

Answer

The required answer is  $7{\left(5x^4-3x^2\right)}^6\left(20x^3-6x\right)\left(2x^3+1\right)+6x^2{\left(5x^4-3x^2\right)}^7$

Step 1. Given Information

The given function is  $f\left(x\right)={\left(5x^4-3x^2\right)}^7\left(2x^3+1\right)$

Step 2. Calculation

Differentiate both the sides with respect to x, we get,      

$$\begin{gathered} f\text{'}\left(x\right)=\left(\frac{d}{dx}{\left(5x^4-3x^2\right)}^7\right)\left(2x^3+1\right)+{\left(5x^4-3x^2\right)}^7\left(\frac{d}{dx}\left(2x^3+1\right)\right) \\ =\left(7{\left(5x^4-3x^2\right)}^6\left(\frac{d}{dx}5x^4-\frac{d}{dx}3x^2\right)\right)\left(2x^3+1\right)+6x^2{\left(5x^4-3x^2\right)}^7 \\ =7{\left(5x^4-3x^2\right)}^6\left(20x^3-6x\right)\left(2x^3+1\right)+6x^2{\left(5x^4-3x^2\right)}^7 \end{gathered}$$