Question

7-52: Find the derivative of the function

9. \(f\left( x \right) = \sqrt {5x + 1} \)

Short answer

The derivative of the function is \(f'\left( x \right) = \frac{5}{{2\sqrt {5x + 1} }}\).

Step-by-step solution

The Power rule combined with the Chain Rule

Power rule is defined as:

\(\frac{d}{{dx}}\left( {{u^n}} \right) = n{u^{n - 1}}\frac{{du}}{{dx}}\)

Additionally, \(\frac{d}{{dx}}{\left( {g\left( x \right)} \right)^n} = n{\left( {g\left( x \right)} \right)^{n - 1}} \cdot g'\left( x \right)\).

Find the derivative of the function

Rewrite the function as the power of \(x\) as shown below:

\(\begin{aligned}f\left( x \right) &= \sqrt {5x + 1} \\ &= {\left( {5x + 1} \right)^{\frac{1}{2}}}\end{aligned}\)

Use the power rule combined with the chain rule to obtain the derivative of the function as shown below:

\(\begin{aligned}f'\left( x \right) &= \frac{d}{{dx}}{\left( {5x + 1} \right)^{\frac{1}{2}}}\\ &= \frac{1}{2}{\left( {5x + 1} \right)^{ - \frac{1}{2}}} \cdot \frac{d}{{dx}}\left( {5x + 1} \right)\\ &= \frac{1}{2}{\left( {5x + 1} \right)^{ - \frac{1}{2}}}\left( 5 \right)\\ &= \frac{5}{{2\sqrt {5x + 1} }}\end{aligned}\)

Thus, the derivative of the function is \(f'\left( x \right) = \frac{5}{{2\sqrt {5x + 1} }}\).