find $\frac{d}{dx}left(\frac{7x^{2}}{3}-\frac{3}{8x^{2}}\right)$. $\frac{d}{dx}left(\frac{7x^{2}}{3}-\frac{3}…

Explanation:

Step1: Use sum - difference rule of derivatives

The derivative of a sum/difference of functions is the sum/difference of their derivatives. So, $\frac{d}{dx}\left(\frac{7x^{2}}{3}-\frac{3}{8x^{2}}\right)=\frac{d}{dx}\left(\frac{7x^{2}}{3}\right)-\frac{d}{dx}\left(\frac{3}{8x^{2}}\right)$.

Step2: Rewrite the second - term

Rewrite $\frac{3}{8x^{2}}$ as $\frac{3}{8}x^{- 2}$. Then our expression becomes $\frac{d}{dx}\left(\frac{7x^{2}}{3}\right)-\frac{d}{dx}\left(\frac{3}{8}x^{-2}\right)$.

Step3: Use the constant - multiple rule of derivatives

The constant - multiple rule states that $\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x))$. So we have $\frac{7}{3}\frac{d}{dx}(x^{2})-\frac{3}{8}\frac{d}{dx}(x^{-2})$.

Step4: Use the power rule of derivatives

The power rule is $\frac{d}{dx}(x^{n})=nx^{n - 1}$. For $\frac{7}{3}\frac{d}{dx}(x^{2})$, we get $\frac{7}{3}\times2x^{2 - 1}=\frac{14}{3}x$. For $\frac{3}{8}\frac{d}{dx}(x^{-2})$, we get $\frac{3}{8}\times(-2)x^{-2 - 1}=-\frac{3}{4}x^{-3}$.

Step5: Combine the results

$\frac{14}{3}x-\left(-\frac{3}{4}x^{-3}\right)=\frac{14}{3}x+\frac{3}{4x^{3}}$.

Answer:

$\frac{14}{3}x+\frac{3}{4x^{3}}$