Thread: Derivative

Find the 2nd derivative of tan(2x).

The first is sec^2(2x). Then you would change it to 1/(cos^2(2x)). How would you go from there? Thanks

Originally Posted by classicstrings

Find the 2nd derivative of tan(2x).

The first is sec^2(2x). Then you would change it to 1/(cos^2(2x)). How would you go from there? Thanks

If you already know how to find the derivative of sec(2x), you don't need to change sec^2(2x) into 1/cos^2(2x).

I will go about this assuming you know how to take the derivative of secant.

f(x) = tan(2x)

The derivative of f(x) involves the chain rule (we need to use it twice):
f'(x) = sec^2(2x)*2 <-- the derivative of tan(2x) times the derivative of 2x.
f'(x) = 2sec^2(2x)

The derivative of f'(x) involves the chain rule (we need to use it three times):
f''(x) = 2*2sec(2x)*sec(2x)tan(2x)*2 <-- 2 times the derivatives of sec^2(2x) times the derivative of sec(2x) times the derivative of 2x.
f''(x) = 8sec^2(2x)tan(2x)

Originally Posted by classicstrings

Find the 2nd derivative of tan(2x).

The first is sec^2(2x). Then you would change it to 1/(cos^2(2x)). How would you go from there? Thanks

No the first is,
2*sec^2(2x)
By chain rule.

But you do not need to change it.

You can write, by Pythagoren identity,

2*(1+tan^2(2x)) = 2 + 2*tan^2(2x)

You you can differenciate.