Q. 26 - 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.   

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

Step-by-Step Solution

Answer

The required answer is $\frac{3}{\sqrt{x^2+1}}-\frac{x(3x+1)}{{\left(\sqrt{x^2+1}\right)}^3}$

Step 1. Given Information

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

Step 2. Calculation

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

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