Q. 45 - 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. 

$f\left(x\right)={\left(x^4-\sqrt{3-4x}\right)}^8+5x$ 

Step-by-Step Solution

Answer

The required answer is $8{\left(x^4-\sqrt{3-4x}\right)}^7\left(4x^3+2{(3-4x)}^{-\frac{3}{2}}\right)+5$

Step 1. Given Information

The given function is  $f\left(x\right)={\left(x^4-\sqrt{3-4x}\right)}^8+5x$

Step 2. Calculation

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

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\left(x^4-\sqrt{3-4x}\right)}^8+\frac{d}{dx}\left(5x\right) \\ =8{\left(x^4-\sqrt{3-4x}\right)}^{8-1}\frac{d}{dx}\left(x^4-\sqrt{3-4x}\right)+5 \\ =8{\left(x^4-\sqrt{3-4x}\right)}^7\left(4x^3-\frac{1}{2}{\left(3-4x\right)}^{-\frac{1}{2}-1}\frac{d}{dx}(3-4x)\right)+5 \\ =8{\left(x^4-\sqrt{3-4x}\right)}^7\left(4x^3-\frac{1}{2}{\left(3-4x\right)}^{-\frac{3}{2}}(-4)\right)+5 \\ =8{\left(x^4-\sqrt{3-4x}\right)}^7\left(4x^3+\frac{4}{2}{\left(3-4x\right)}^{-\frac{3}{2}}\right)+5 \\ =8{\left(x^4-\sqrt{3-4x}\right)}^7\left(4x^3+2{\left(3-4x\right)}^{-\frac{3}{2}}\right)+5 \end{gathered}$$