Derivatives?

Question

I have tried time and again to differentiate the following expression, but cannot seem to get it. Help please!

x^(x cos(x))

Thanks!

sorry, it's actually:

x^ (cos(x))

Answer 1

anytime you encounter something like f(x)^g(x), you have to do the same thing:

see it as e^(g(x)*ln(f(x)) (check, and you'll see this is the same as f(x)^(gx))

x^(xcosx) = e^((xcosx)lnx)

the derivative = e^(xcosxlnx) * the derivative of the exponent (chain rule)

derivative of the exponent = cosxlnx - xsinxlnx + cosx

so the entire derative is (cosxlnx - xsinxlnx + cosx)*x^(xcosx)

Answer 2

First, rewrite the function like this:
x^(x cos(x)) = e ^ (ln(x) cos(x))

Then, differentiate the form that you see on the right. You will need to use the chain rule, and the product rule.

The derivative is
e ^ (ln(x) * cos(x)) (-ln(x) sin(x) + cos(x)/x)

Answer 3

That's tedious but not tough.

You need to repeatedly use the chain rule. It might help to substitute some other symbols for bookeeping

First,

f(x) = x^g(x) now use the chain rule

for the g'(x) part use the chain rule on
g(x)=xcos(x)

Making sure to use lots of ( ) and [ ] and even { }, if you need them!

Answer 4

Let's see:
[x^(xcos(x))]'
when i got this form such as: x^n
I can (x^n)' by nx'x^n-1 when n is a real and lnx when n is natural.we got too;
(xy)'=x'y+y'x when xy are natural.And then, (xcosx)'=x'cosx-xsinx
For [x^(xcos(x)]' =(x'cosx-xsinx)lnx
with x'=1
'' =lnx^(cosx-xsinx)
or
[x^(xcos(x)]' =ln^cosx-lnsinx^x