Deegan Chase

2022-03-22

Find the derivatives of the functions
$z=\frac{4-3x}{3{x}^{2}+x}$

### Answer & Explanation

Brennan Thompson

Step 1
The derivative of a function is the rate of change of the function with respect to the function variable. The derivative of a function of the form $\frac{f\left(x\right)}{g\left(x\right)}$ can be calculated using the quotient rule, which is given by,
${\left(\frac{f\left(x\right)}{g\left(x\right)}\right)}^{\prime }=\frac{g\left(x\right)\cdot {f}^{\prime }\left(x\right)-f\left(x\right)\cdot {g}^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^{2}}$
Step 2
Therefore, the derivative of the function $z=\frac{4-3x}{3{x}^{2}+x}$ is given by,
${z}^{\prime }=\frac{d\left(\frac{4-3x}{3{x}^{2}+x}\right)}{dx}$
$=\frac{\left(3{x}^{2}+x\right)\cdot \left(\frac{d\left(4-3x\right)}{dx}\right)-\left(4-3x\right)\cdot \left(\frac{d\left(3{x}^{2}+x\right)}{dx}\right)}{{\left(3{x}^{2}+x\right)}^{2}}$
$=\frac{\left(3{x}^{2}+x\right)\cdot \left(-3\right)-\left(4-3x\right)\cdot \left(6x+1\right)}{{\left(3{x}^{2}+x\right)}^{2}}$
$=\frac{-9{x}^{2}-3x-\left(24x+4-18{x}^{2}-3x\right)}{{\left(3{x}^{2}+x\right)}^{2}}$
$=\frac{-9{x}^{2}-3x-24x-4+18{x}^{2}+3x}{{\left(3{x}^{2}+x\right)}^{2}}$
$=\frac{9{x}^{2}-24x-4}{{\left(3{x}^{2}+x\right)}^{2}}$
Therefore, the derivative of the function is $\frac{9{x}^{2}-24x-4}{{\left(3{x}^{2}+x\right)}^{2}}$

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