Deegan Chase

2022-03-22

Find the derivatives of the functions

$z=\frac{4-3x}{3{x}^{2}+x}$

Brennan Thompson

Beginner2022-03-23Added 10 answers

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,

$z}^{\prime}=\frac{d\left(\frac{4-3x}{3{x}^{2}+x}\right)}{dx$

$=\frac{(3{x}^{2}+x)\cdot \left(\frac{d(4-3x)}{dx}\right)-(4-3x)\cdot \left(\frac{d(3{x}^{2}+x)}{dx}\right)}{{(3{x}^{2}+x)}^{2}}$

$=\frac{(3{x}^{2}+x)\cdot (-3)-(4-3x)\cdot (6x+1)}{{(3{x}^{2}+x)}^{2}}$

$=\frac{-9{x}^{2}-3x-(24x+4-18{x}^{2}-3x)}{{(3{x}^{2}+x)}^{2}}$

$=\frac{-9{x}^{2}-3x-24x-4+18{x}^{2}+3x}{{(3{x}^{2}+x)}^{2}}$

$=\frac{9{x}^{2}-24x-4}{{(3{x}^{2}+x)}^{2}}$

Therefore, the derivative of the function is$\frac{9{x}^{2}-24x-4}{{(3{x}^{2}+x)}^{2}}$

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

Step 2

Therefore, the derivative of the function

Therefore, the derivative of the function is

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