MHB What is the 2nd Derivative of y(t)=tan5t?

[riri]

Hello!

I'm trying to find the 2nd derivative of y(t)=tan5t.
I first found the first derivative.. and got y'(t)=sec^2(5t)(5) --> 5sec^2(5t)
--> 5/(cos^2(5t)

But to find the 2nd derivative I'm confused...
I got until y"(t)=\frac{cos^2(5t)(5)'-(5)(cos^2(5t))'}{(cos^2(5t)(cos^2(5t))}

 

[Prove It]

Hello!

I'm trying to find the 2nd derivative of y(t)=tan5t.
I first found the first derivative.. and got y'(t)=sec^2(5t)(5) --> 5sec^2(5t)
--> 5/(cos^2(5t)

But to find the 2nd derivative I'm confused...
I got until y"(t)=\frac{cos^2(5t)(5)'-(5)(cos^2(5t))'}{(cos^2(5t)(cos^2(5t))}

Write it as $\displaystyle \begin{align*} y'(t) = 5\left[ cos{(5t)} \right] ^{-2} \end{align*}$ and apply the Chain Rule.