How would I calculate this second derivative?

[bronze cobalt]

chain rule twice

[lean robin]

I will write down my steps and show you what I get. Is that ok?

[bronze cobalt]

that's fine

[lean robin]

This is what I have, i'm not sure how to simplify it now tho. Do you have a clue?

[bronze cobalt]

why do you need to simplify?

[lean robin]

It's required and will help me for the next part of this question

[bronze cobalt]

well you would get sin^2(5x) - cos^2(5x)

times 50 which would simplify to -50cos(10x)

you're final answer for the derivative is correct thouhg, mind sending the next part?

[lean robin]

I just wanted to confirm if i'm on the right track here

I was able to find g'(x) and g''(x)
But i'm confused when it says line tangent to g'(x)
Does this mean my tangent line equation will be
y = g'(a) + g''(a) (x - a) with a = pi ?

[bronze cobalt]

you don't need to simplify, you know the values of sin^2(5pi), cos^2(5pi), etc.

its always going to be either 0 or 1!

but yes you are on the right track

[lean robin]

ok

cuz usually the tangent line equation is simply y = g(a) + g'(a) (x - a)

but since this question says tanget to g'

It would now become:

y = g'(a) + g''(a) (x - a) with a = pi , right?

[bronze cobalt]

yep

[lean robin]

ok cool

Thanks a lot!

[bronze cobalt]

no problem, if you have any other questions ask them, or else do .close

[lean robin]

yes I had another question

When I have a tangent line equation:
y = f(a) + f'(a) (x - a)
f'(a) is the slope right?

[bronze cobalt]

yep

[lean robin]

ok cool

One last question

For part a) here: if I plug in cosh^-1(x) into cosh(x) and simply that to just x, is that all I need to do to prove it is the inverse because if we have a function f(x) the inverse function is g(x) if f(g(x)) = x?

[bronze cobalt]

that would wrok

but the easier way would be to swap x and y for the definition of cosh

[lean robin]

ok ya

but my method is still correct right?

[bronze cobalt]

yep

[lean robin]

ok nice

[arctic phoenix]

.solved