Question

find $\frac{d}{dx}left(\frac{2x^{2}}{3}-\frac{7}{4x^{2}}
ight)$. $\frac{d}{dx}left(\frac{2x^{2}}{3}-\frac{7}{4x^{2}}
ight)=square$

Explanation:

Step1: Use sum - difference rule of differentiation

The derivative of a sum or difference of functions \(y = u - v\) is \(y'=u' - v'\). Let \(u=\frac{2x^{2}}{3}\) and \(v = \frac{7}{4x^{2}}\).

Step2: Differentiate \(u=\frac{2x^{2}}{3}\)

Using the power - rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), for \(u=\frac{2}{3}x^{2}\), we have \(u'=\frac{2}{3}\times2x^{2 - 1}=\frac{4x}{3}\).

Step3: Rewrite \(v=\frac{7}{4x^{2}}\) and differentiate

Rewrite \(v=\frac{7}{4}x^{- 2}\). Then, by the power - rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), \(v'=\frac{7}{4}\times(-2)x^{-2 - 1}=-\frac{7}{2}x^{-3}=-\frac{7}{2x^{3}}\).

Step4: Calculate the derivative of the whole expression

\(\frac{d}{dx}(\frac{2x^{2}}{3}-\frac{7}{4x^{2}})=u' - v'=\frac{4x}{3}+\frac{7}{2x^{3}}\).

Answer:

\(\frac{4x}{3}+\frac{7}{2x^{3}}\)