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.

$f\left(x\right)={\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{100}$

Short answer

The required answer is$100{\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{99}\left(180x^3{\left(3x^4-1\right)}^2+3\right)$

Step-by-step solution

Step 1. Given Information  

The given function is $f\left(x\right)={\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{100}$

Step 2. Calculation   

Differentiate both the sides with respect tox, we get,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{100} \\ =100{\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{100-1}\left(\frac{d}{dx}\left(5{\left(3x^4-1\right)}^3\right)+\frac{d}{dx}\left(3x\right)-\frac{d}{dx}1\right) \\ =100{\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{99}\left(5\left(3{\left(3x^4-1\right)}^{3-1}\right)\frac{d}{dx}(3x^4-1)+3\right) \\ =100{\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{99}\left(15{\left(3x^4-1\right)}^2\left)\right(12x^3)+3\right) \\ =100{\left(5{\left(3x^4-1\right)}^3+3x-1\right)}^{99}\left(180x^3{\left(3x^4-1\right)}^2+3\right) \end{gathered}$$