Calculus problem, derivative

[bobandjohn]

find the derivative of y = cos ( X ^ (root X))

i think you have to use the chain rule but i dont know how please help thanks

[HallsofIvy]

Okay, the chain rule SAYS f(g(x))'= f'(g(x))g'(x). The nice thing about it is that it allows you to work with one function at a time!
Here y= cos(x^sqrt(x))

Okay, y= cos(u) where u= x^sqrt(x).
Differentiating cos(u) (with respect to u) is easy: -sin(u)= -sin(x^sqrt(x)).

Now, you just have to find the derivative of u= x^sqrt(x) and multiply by it. I would recommend "logarithmic differentiation: take the ln of both sides to get
ln(u)= ln(x^sqrt(x))= sqrt(x) ln(x). The derivative of ln(u) is (1/u)u' and you will need to use the product rule for sqrt(x)ln(x).

[bobandjohn]

hey thanks for the reply

i get what you are saying
but how do you just frind the derivative of u = X^ sqrt(x)
is the log rule the only way ?

[HallsofIvy]

Any time you have a function of x to the power a function of x, f(x)^g(x), logarithmic differentiation is easiest. No, it is not necessary. In fact, there is a 'simple' formula (derived by logarithmic differentiation):
(f(x)^g(x))'= g(x)f(x)^(g(x)-1)f'(x)+ f(x)^g(x) ln(f(x)) g'(x). I certainly wouldn't recommend trying to memorize that!

For something like y= x^sqrt(x), logarithmic differentiation is simple:
ln(y)= sqrt(x) ln(x)= x^(1/2) ln(x). Differentiating with respect to x on both sides, (ln(y))'= (1/y)y'= (1/2)x^(-1/2)ln(x)+ x^(1/2)/x.
y'= [(1/2)x^(-1/2)ln(x)+ x^(-1/2)](x^sqrt(x))