Question

find the derivative.
$\frac{d}{dt}4(t^{2}+5t)^{-6}$
$\frac{d}{dt}4(t^{2}+5t)^{-6}=square$

Explanation:

Step1: Factor out the constant

Since the derivative of a constant - multiple of a function is the constant times the derivative of the function, and we have $y = 4(t^{2}+5t)^{-6}$, then $\frac{d}{dt}(4(t^{2}+5t)^{-6})=4\frac{d}{dt}(t^{2}+5t)^{-6}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{-6}$ and $u=t^{2}+5t$, then $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$: $\frac{dy}{du}=-6u^{-7}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n=-6$). Second, find $\frac{du}{dt}$: $\frac{du}{dt}=2t + 5$ (using the power rule for $t^{2}$ and $5t$).

Step3: Substitute and simplify

$\frac{dy}{dt}=4\cdot(-6u^{-7})\cdot(2t + 5)$. Substitute $u=t^{2}+5t$ back in: $\frac{dy}{dt}=-24(t^{2}+5t)^{-7}(2t + 5)=-\frac{24(2t + 5)}{(t^{2}+5t)^{7}}$.

Answer:

$-\frac{24(2t + 5)}{(t^{2}+5t)^{7}}$