nicekikah

## Answered question

2021-02-16

Find the derivatives.
$f\left(x\right)=\frac{3{x}^{3}-4}{2{x}^{2}+3}$

### Answer & Explanation

estenutC

Skilled2021-02-18Added 81 answers

Use quotient rule of derivative and find the derivatives of above function.
$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-{g}^{\prime }\left(x\right)f\left(x\right)}{{\left[g\left(x\right)\right]}^{2}}$ [Quotient rule]
$f\left(x\right)=\frac{3{x}^{3}-4}{2{x}^{2}+3}$
${f}^{\prime }\left(x\right)=\frac{\left(2{x}^{2}+3\right)\frac{d}{dx}\left(3{x}^{3}-4\right)-\left(3{x}^{3}-4\right)\frac{d}{dx}\left(2{x}^{2}+3\right)}{{\left(2{x}^{2}+3\right)}^{2}}$
$=\frac{\left(2{x}^{2}+3\right)\left(9{x}^{2}\right)-\left(3{x}^{3}-4\right)\left(4x\right)}{{\left(2{x}^{2}+3\right)}^{2}}$
$=\frac{18{x}^{4}+27{x}^{2}-12{x}^{4}+16x}{{\left(2{x}^{2}+3\right)}^{2}}$
$=\frac{6{x}^{4}+27{x}^{2}+16x}{{\left(2{x}^{2}+3\right)}^{2}}$

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