Question

find the derivative of y with respect to s.

$y = sec^{-1}(5s^{5}+2)$

$\frac{dy}{ds}=square$

Explanation:

Step1: Recall the derivative formula for $y = \sec^{-1}(u)$

The derivative of $y=\sec^{-1}(u)$ with respect to $u$ is $\frac{1}{|u|\sqrt{u^{2}-1}}\frac{du}{ds}$. Here $u = 5s^{5}+2$.

Step2: Find the derivative of $u$ with respect to $s$

Differentiate $u = 5s^{5}+2$ with respect to $s$. Using the power - rule $\frac{d}{ds}(as^{n})=ans^{n - 1}$, we get $\frac{du}{ds}=25s^{4}$.

Step3: Apply the chain - rule

By the chain - rule $\frac{dy}{ds}=\frac{1}{|u|\sqrt{u^{2}-1}}\cdot\frac{du}{ds}$. Substitute $u = 5s^{5}+2$ and $\frac{du}{ds}=25s^{4}$ into the formula.
So $\frac{dy}{ds}=\frac{25s^{4}}{|5s^{5}+2|\sqrt{(5s^{5}+2)^{2}-1}}$.

Answer:

$\frac{25s^{4}}{|5s^{5}+2|\sqrt{(5s^{5}+2)^{2}-1}}$