Question

3-34: Differentiate the function.

17. \(g\left( x \right) = \frac{1}{{\sqrt x }} + \sqrt[4]{x}\)

Short answer

The derivative of the function is \(g'\left( x \right) = - \frac{1}{{2x\sqrt x }} + \frac{1}{{4\sqrt[4]{{{x^3}}}}}\).

Step-by-step solution

Differentiation rule

When \(n\) is any real number, then;

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

When \(f\) and \(g\) are both differentiable, then;

\(\begin{aligned}\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) &= \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\\\frac{d}{{dx}}\left( {f\left( x \right) - g\left( x \right)} \right) &= \frac{d}{{dx}}f\left( x \right) - \frac{d}{{dx}}g\left( x \right)\end{aligned}\)

Differentiate the function

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

\(g\left( x \right) = {x^{ - \frac{1}{2}}} + {x^{\frac{1}{4}}}\)

Differentiate the function as shown below:

\(\begin{aligned}g'\left( x \right) &= \frac{d}{{dx}}\left( {{x^{ - \frac{1}{2}}} + {x^{\frac{1}{4}}}} \right)\\ &= \frac{d}{{dx}}\left( {{x^{ - \frac{1}{2}}}} \right) + \frac{d}{{dx}}\left( {{x^{\frac{1}{4}}}} \right)\\ &= - \frac{1}{2}{x^{ - \frac{1}{2} - 1}} + \frac{1}{4}{x^{\frac{1}{4} - 1}}\\ &= - \frac{1}{2}{x^{ - \frac{3}{2}}} + \frac{1}{4}{x^{\frac{{ - 3}}{4}}}\\ &= - \frac{1}{{2x\sqrt x }} + \frac{1}{{4\sqrt[4]{{{x^3}}}}}\end{aligned}\)

Thus, the derivative of the function is \(g'\left( x \right) = - \frac{1}{{2x\sqrt x }} + \frac{1}{{4\sqrt[4]{{{x^3}}}}}\).