Question

19 mark for review if $y = \frac{1}{2}x^{4/3}-\frac{3}{x^{5}}$, then $\frac{dy}{dx}=$
a $\frac{2}{3x^{1/3}}+\frac{15}{x^{6}}$
b $\frac{2}{3x^{1/3}}+\frac{15}{x^{5}}$
c $\frac{2}{3x^{1/3}}-\frac{3}{5x^{4}}$
d $\frac{2x^{1/3}}{3}+\frac{15}{x^{6}}$
e $\frac{2x^{1/3}}{3}-\frac{3}{5x^{4}}$

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $\frac{dy}{dx}=anx^{n - 1}$. For the first term $y_1=\frac{1}{2}x^{4/3}$, where $a=\frac{1}{2}$ and $n = \frac{4}{3}$. So, $\frac{dy_1}{dx}=\frac{1}{2}\times\frac{4}{3}x^{\frac{4}{3}-1}=\frac{2}{3}x^{\frac{1}{3}}=\frac{2}{3x^{- \frac{1}{3}}}$. For the second term $y_2 =-\frac{3}{x^5}=-3x^{-5}$, where $a=-3$ and $n=-5$. Then $\frac{dy_2}{dx}=(-3)\times(-5)x^{-5 - 1}=15x^{-6}=\frac{15}{x^{6}}$.

Step2: Combine the derivatives

Since $y = y_1 + y_2$, by the sum - rule of differentiation $\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}=\frac{2}{3x^{\frac{1}{3}}}+\frac{15}{x^{6}}$.

Answer:

A. $\frac{2}{3x^{\frac{1}{3}}}+\frac{15}{x^{6}}$