Find the derivative of h(x)=(4x^3-7x+8)/x

Find the derivative of $h=\left(x\right)=\frac{4{x}^{3}-7x+8}{x}$
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Colin Dougherty
Step 1
Given: $h=\left(x\right)=\frac{4{x}^{3}-7x+8}{x}$
To find- The derivative of the above function.
Identity Used- Using the derivative of the quotient function as
$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{g\left(x\right)×{f}^{\prime }\left(x\right)-f\left(x\right)×{g}^{\prime }\left(x\right)}{g\left(x{\right)}^{2}}$, where ,br.f(x) and g(x) are the function of x.
Step 2
Explanation- Rewrite the given expression,
$h\left(x\right)=\frac{4{x}^{3}-7x+8}{x}$
Using the derivative of the quotient function, differentiating the above expression w.r.t. x, we get,
${h}^{\prime }\left(x\right)=\frac{x\left(12{x}^{2}-7\right)-\left(4{x}^{3}-7x+8\right)\cdot 1}{{x}^{2}}$
$=\frac{12{x}^{3}-7x-4{x}^{3}+7x-8}{{x}^{2}}$
$=\frac{8{x}^{3}-8}{{x}^{2}}$
$=\frac{8\left({x}^{3}-1\right)}{{x}^{2}}$
So, the derivative of the above function is
${h}^{\prime }\left(x\right)=\frac{8\left({x}^{3}-1\right)}{{x}^{2}}$
Answer- the derivative of the function
$\frac{4{x}^{3}-7x+8}{x}$ is ${h}^{\prime }\left(x\right)=\frac{8\left({x}^{3}-1\right)}{{x}^{2}}$