Question

find y. y = \frac{1}{36} \tan(6x + 5) y = \square

Explanation:

Step1: Find first - derivative using chain - rule

The derivative of $\tan(u)$ is $\sec^{2}(u)$ and if $y = \frac{1}{36}\tan(6x + 5)$ with $u = 6x+5$, then $y'=\frac{1}{36}\sec^{2}(6x + 5)\cdot6=\frac{1}{6}\sec^{2}(6x + 5)$.

Step2: Find second - derivative using chain - rule and product - rule

We can write $y'=\frac{1}{6}\sec^{2}(6x + 5)=\frac{1}{6}(\sec(6x + 5))^{2}$. Let $v=\sec(6x + 5)$, so $y'=\frac{1}{6}v^{2}$.
First, find the derivative of $v=\sec(6x + 5)$. The derivative of $\sec(u)$ is $\sec(u)\tan(u)$, and with $u = 6x+5$, $v'=\sec(6x + 5)\tan(6x + 5)\cdot6 = 6\sec(6x + 5)\tan(6x + 5)$.
By the chain - rule, $y''=\frac{1}{6}\cdot2v\cdot v'$. Substitute $v=\sec(6x + 5)$ and $v' = 6\sec(6x + 5)\tan(6x + 5)$ into the formula for $y''$.
$y''=\frac{1}{6}\cdot2\sec(6x + 5)\cdot6\sec(6x + 5)\tan(6x + 5)=2\sec^{2}(6x + 5)\tan(6x + 5)$.

Answer:

$2\sec^{2}(6x + 5)\tan(6x + 5)$