find f'(x)
(sinx)^cosx
See the generalized power rule here:
http://en.wikipedia.org/wiki/Table_of_derivatives
Use the chain rule for differentiating a function of a function.
Let cosx = v(x)
d/dx[(sinx)^v(x)]= d(sinx)^v/dv * dv/dx
= d/dv[sqrt(1-v^2)]/dv * dv/dx
= (-2)(1/2)/[sqrt(1-v^2)] * (-sin x)
= (-1)/sin x * (-sin x) = 1
A very interesting result!
My answer looks rather suspicious, although I verified it at x = 0 and x = 0.5. I hope that other teachers will respond.
I would do it this way:
take ln of both sides
lny = ln(cosx^sinx)
= sinx(ln(cosx))
(dy/dx)/y = sinx(cosx/sinx) + (-sinx)(ln(sinx)
dy/dx = (cosx)^sinx[sinx(cosx/sinx) + (-sinx)(ln(sinx)]
don't know how much simplication is needed.
Got my sines and cosines mixed up as I copied from paper.
try again:
lny = ln(sinx^cosx)
= cosx(ln(sinx))
(dy/dx)/y = cosx(cosx/sinx) + (-sinx)(ln(sinx))
dy/dx = dy/dx = (sinx)^cosx[cosx(cosx/sinx) + (-sinx)(ln(sinx)]
To find the derivative of the function f(x) = (sin(x))^cos(x), we can use the chain rule.
Let's break down the function into two parts:
u = sin(x)
v = cos(x)
Now, we can rewrite the function as f(x) = u^v.
To find f'(x), the derivative of f(x) with respect to x, we use the chain rule:
f'(x) = v * u^(v-1) * u' + u^v * ln(u) * v'
Applying this to our function, we have:
f'(x) = cos(x) * (sin(x))^(cos(x)-1) * cos(x) + (sin(x))^cos(x) * ln(sin(x)) * (-sin(x))
Simplifying further, we get:
f'(x) = cos(x) * (sin(x))^(cos(x)-1) * cos(x) - (sin(x))^cos(x) * ln(sin(x)) * sin(x)
So, the derivative of f(x) = (sin(x))^cos(x) is given by f'(x) = cos(x) * (sin(x))^(cos(x)-1) * cos(x) - (sin(x))^cos(x) * ln(sin(x)) * sin(x).