Question

3-34 Differentiate the function.

15. \(y = 2x + \sqrt x \)

Short answer

The differentiation of the function \(y = 2x + \sqrt x \) is \(y' = 2 + \frac{1}{{2\sqrt x }}\).

Step-by-step solution

Write the formula of the sum rules, the power function and the power rule

The Sum Rule:\(\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\)

The Power function:\(\frac{d}{{dx}}\left( x \right) = 1\)

The Power Rule: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\) where \(n\) is a positive integer

Find the differentiation of the function

Consider the function \(y = 2x + \sqrt x \). Differentiate the function w.r.t \(x\) by using the sum rule, the power function and the power rule.

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

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