Q. 42 - 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.  

$f\left(x\right)=x\sqrt{3x^2+1}\sqrt[3]{2x+5}$

Step-by-Step Solution

Answer

The required answer is $\sqrt{3x^2+1}\sqrt[3]{2x+5}+\frac{3x^2\sqrt[3]{2x+5}}{\sqrt{3x^2+1}}+3x\sqrt{3x^2+1}\sqrt[3]{2x+5}$

Step 1. Given Information

The given function is   $f\left(x\right)=x\sqrt{3x^2+1}\sqrt[3]{2x+5}$

Step 2. Calculation

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

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