Question

If $y=t^{\frac{4}{3}}-3t^{\frac{-2}{3}}$, then $\frac{dy}{dt}=$

• A. $\frac{2t^2+3}{3t^{{\displaystyle \frac{5}{3}}}}$
• B. $\frac{2t^2+3}{t^{{\displaystyle \frac{5}{3}}}}$
• C. $\frac{2(2t^2+3)}{t^{{\displaystyle \frac{5}{3}}}}$
• D. $\frac{2(2t^2+3)}{3t^{{\displaystyle \frac{5}{3}}}}$

Answer 1

The correct option is D

$\frac{2(2t^2+3)}{3t^{{\displaystyle \frac{5}{3}}}}$

Explanation for the correct option:

$y=t^{\frac{4}{3}}-3t^{\frac{-2}{3}}$

Differentiating with respect to $x$ both sides:

$$\begin{gathered} \frac{dy}{dx}=\frac{4}{3}{t}^{\frac{4}{3}-1}-3\times \frac{\left(-2\right)}{3}{t}^{\frac{-2}{3}-1} \\ \Rightarrow \frac{dy}{dx}=\frac{4}{3}{t}^{\frac{1}{3}}+2t^{\frac{-5}{3}} \\ \Rightarrow \frac{dy}{dx}=\frac{4}{3}{t}^{\frac{1}{3}}+\frac{2}{t^{\frac{5}{3}}} \\ \Rightarrow \frac{dy}{dx}=\frac{4t^{{\displaystyle \frac{1}{3}}}\times t^{\frac{5}{3}}+6}{3t^{\frac{5}{3}}} \\ \Rightarrow \frac{dy}{dx}=\frac{4t^2+6}{3t^{\frac{5}{3}}} \\ \Rightarrow \frac{dy}{dx}=\frac{2(2t^2+3)}{3t^{\frac{5}{3}}} \end{gathered}$$

Hence, Option (D) is the correct answer.