Q. 50 - Calculus

Question

Use the differentiation rules developed in this section to find the derivatives of the functions in Exercises 35-64. Note that it may be necessary to do some preliminary algebra before differentiating.    

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

Step-by-Step Solution

Answer

The derivative of the function is $f\text{'}\left(x\right)=\begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{5}{3}\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & \frac{-1}{6}\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \frac{2}{3}\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & \frac{-1}{3}\end{array}$.

Step 1. Given Information

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

Step 2. Simplify the function

Apply the identity ${(a-b)}^2=a^2-2ab+b^2$ in the given function.

$\begin{array}{rcl}f\left(x\right)& =& {\left(\sqrt{x}\right)}^2-2\sqrt{x}\sqrt[3]{x}+{\left(\sqrt[3]{x}\right)}^2 =x-2x^{\frac{1}{2}+\frac{1}{3}}+x^{\frac{2}{3}}\\ & =& x-2x^{\frac{5}{6}}+x^{\frac{2}{3}}\end{array}$

Step 3. Find the derivative

• Apply the sum rule of derivative, $(f+g)\text{'}\left(x\right)=f\text{'}\left(x\right)+g\text{'}\left(x\right)$ and the difference rule of derivative, $(f-g)\text{'}\left(x\right)=f\text{'}\left(x\right)-g\text{'}\left(x\right)$.

$f\text{'}\left(x\right)=\frac{d}{dx}(\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & )\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{d}{dx}\end{array}\begin{array}{rcl}& & (\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^{\frac{5}{6}}\end{array}\begin{array}{rcl}& & )\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \frac{d}{dx}\end{array}\begin{array}{rcl}& & (\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^{\frac{2}{3}}\end{array}\begin{array}{rcl}& & )\end{array}$

• Apply the constant multiple rule of derivative, $\left(kf\right)\text{'}\left(x\right)=kf\text{'}\left(x\right)$.

$f\text{'}\left(x\right)=\frac{d}{dx}(\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & )\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\frac{d}{dx}\end{array}\begin{array}{rcl}& & (\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^{\frac{5}{6}}\end{array}\begin{array}{rcl}& & )\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \frac{d}{dx}\end{array}\begin{array}{rcl}& & (\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^{\frac{2}{3}}\end{array}\begin{array}{rcl}& & )\end{array}$

• Apply the power rule of derivative, $\left(x^n\right)\text{'}=n{x}^{n-1}$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \left(1\right)-2\left(\frac{5}{6}{x}^{\frac{-1}{6}}\right)+\left(\frac{2}{3}{x}^{\frac{-1}{3}}\right)\\ & =& 1-\frac{5}{3}{x}^{\frac{-1}{6}}+\frac{2}{3}{x}^{\frac{-1}{3}}\end{array}$