Simple question

Derivative of \( \boxed{f(y)= \frac{y^2}{y^3+8} } \)

Let’s go ; D

\( f(y)=\frac{y^2}{y^3+8} \)

we have to use the quotient rule.

\( f(y)=\frac{g(y)}{h(y)} \)

\( f’(y)=\frac{h(y)*g’(y)-g(y)*h’(y)}{[h(y)]^2} \)

Then

\( g(y)=y^2 \)

\( g’(y)=2y \)

\( h(y)=y^3+8 \)

\( h(y)=3y^2 \)

Now we can replace

\( f’(y)=\frac{h(y)*g’(y)-g(y)*h’(y)}{[h(y)]^2} \)

\( f’(y)=\frac{(y^3+8)*2y-(y^2)*3y^2}{(y^3+8)^2} \)

\( f’(y)=\frac{2y^4+16y-3y^4}{(y^3+8)^2} \)

\( \boxed{\boxed{f’(y)=\frac{16y-y^4}{(y^3+8)^2}}} \)


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