Question

differentiate the function. s(t) = 1/t + 1/t^6 s(t) =

Explanation:

Step1: Rewrite the function

Rewrite $\frac{1}{t}$ as $t^{-1}$ and $\frac{1}{t^{6}}$ as $t^{-6}$, so $s(t)=t^{-1}+t^{-6}$.

Step2: Apply power - rule

The power - rule for differentiation is $\frac{d}{dt}(t^{n})=nt^{n - 1}$.
For the first term $t^{-1}$, its derivative is $(-1)t^{-1 - 1}=-t^{-2}$.
For the second term $t^{-6}$, its derivative is $(-6)t^{-6 - 1}=-6t^{-7}$.

Step3: Combine the derivatives

$s'(t)=-t^{-2}-6t^{-7}$.
We can rewrite it as $s'(t)=-\frac{1}{t^{2}}-\frac{6}{t^{7}}$.

Answer:

$-\frac{1}{t^{2}}-\frac{6}{t^{7}}$