Question

use the quotient rule to find the derivative of the following (y = \frac{7x^{2}+1}{x^{2}+4}) (\frac{dy}{dx}=square)

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 7x^{2}+1$, $v=x^{2}+4$.

Step2: Find $u^\prime$ and $v^\prime$

Differentiate $u = 7x^{2}+1$ with respect to $x$: $u^\prime=\frac{d}{dx}(7x^{2}+1)=14x$. Differentiate $v=x^{2}+4$ with respect to $x$: $v^\prime=\frac{d}{dx}(x^{2}+4)=2x$.

Step3: Apply the quotient - rule

Substitute $u$, $v$, $u^\prime$, and $v^\prime$ into the quotient - rule formula.

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Answer:

$\frac{54x}{(x^{2}+4)^{2}}$