Question

Can someone explain how to differentiate y=cos^2(3x)?

Answer 1

\(\large \cos^2(3x)\) is same as \(\large (\cos(3x))^2\)

Use chain rule for this.

Answer 2

Thank you!

Answer 3

Wait, actually, I'm still a bit confused. Wouly you also take the derivative of (3x)?

Answer 4

*Would

Answer 5

Yes.

Answer 6

So is everything clear?

Answer 7

I think I need to have the steps be broken down, because I keep getting the wrong answer... @geerky42

Answer 8

Do you know how chain rule works?
\[\dfrac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)\] For function that is made of 3 parent functions; \[\dfrac{d}{dx}f(g(h(x))) = f'(g(h(x)))\cdot g'(h(x)) \cdot h'(x)\] Basically, first you derivative "outer layer", then you do "inner", then you derivative inner of inner and so on.

Is that clear?

Answer 9

That's perfect! Thank you! Also I have one more question, if you don't mind.

Answer 10

So you can see, \(\cos^2(3x)\) can be seen as \(f(g(h(x))),\) where \(f(x) = x^2,~g(x)=\cos x,~h(x)=3x\)

Answer 11

Sure

Answer 12

How would I take the derivative of f(x)=e^(2/x)?

Answer 13

Chain rule again, f(x) = e^x and g(x) = 2/x

Answer 14

Oh, okay! Why wouldn't it just be as it is? Because the derivative of e^x is e^x, so I'm a bit confused.

Answer 15

Of course, but you have to derivative inner too. So derivative of e^(2/x) is not e^(2/x)

Answer 16

you would have to take derivative of inner, 2/x, then multiply it to e^(2/x). Does that make sense so far?

Answer 17

Yeah

Answer 18

I got the answer right! Thank you for all the help!

Answer 19

No problem!