Question

Find the derivative of the function

Answer 1

\[y = x^{\cos(x)}\]

I know the property \[f(x) = b^{x} \] \[f'(x) = b^x \ln(b)\]

so I thought it would be \[x^{\cos(x)}-\sin(x)\ln(x)\]

Answer 2

but that is wrong

Answer 3

Well you know power rule
\(\large\rm y=x^c\qquad\to\qquad y'=c x^{c-1}\)

and you know your exponent rule
\(\large\rm y=c^x\qquad\to\qquad y'=c^x\ln c\)

But it seems like we have neither of those going on here, ya? :)
We have function of x as the base AND exponent.

Answer 4

I was fiddling around and found out that when you have this type of thing going on (function to a function),
it ends up being the sum of these two rules!
power rule + exponent rule.

But let's ignore that for now.
They want you to apply `logarithmic differentiation` here :)

Answer 5

okay so I will get \[\ln(y) = \ln(x^{\cos(x)})\]

then have \[\frac{ y' }{ y } = \cos(x)\ln(x)\]

Answer 6

First step looks good.
But it looks like you differentiated the left side and not the right :O weird lol

Answer 7

Right side, we'd set up our product rule, ya?

Answer 8

right, so \[\frac{ y' }{ y } = -\sin(x)\ln(x) + \frac{ 1 }{ x } \cos(x)\]

Answer 9

Mmm k looks great.

Answer 10

I can just leave the y'/y?

Answer 11

No. They want to know the derivative, which is y'=stuff.
So we'll have just a little bit more work from here :)

Answer 12

I guess multiply by y would be a good step, ya?

Answer 13

\[y' = (y)[(-\sin(x)\ln(x) + (1/x)\cos(x))]\]

Answer 14

Our result should be `in terms of x`.
So we'll replace y with its equivalent that we started with.

Answer 15

oh!!

Answer 16

\[y' = (x^{\cos(x)})[(-\sin(x)\ln(x) + \frac{ 1 }{ x }\cos(x))]\]

Answer 17

\[\large\rm y'=x^{\cos x}\left[\frac{1}{x}\cos x-\sin x \ln x\right]\]Ah yes, great job! :)
Looks a little better without so many brackets haha.
Sorry to be so fussy XD

Answer 18

Thanks, completely forgot about that!