Find the derivative of f(x)=[lnx]^4
let z = ln x
dz = dx/x
dx = x dz
then
f(z) = z^4
df = 4 z^3 dz = 4 z^3 dx/x
df/dx = (4 /x) (ln x)^3
Thank you! I've been stuck on that problem all night.
To find the derivative of the function f(x) = [ln(x)]^4, we can use the chain rule.
The chain rule states that if we have a composite function of the form g(h(x)), where g(u) is a differentiable function and h(x) is also differentiable, then the derivative of g(h(x)) is given by the product of the derivative of g(u) with respect to u and the derivative of h(x) with respect to x.
In this case, let g(u) = u^4 and h(x) = ln(x). To find the derivative of f(x), we need to find the derivatives of g(u) and h(x).
The derivative of g(u) = u^4 with respect to u is 4u^(4-1) = 4u^3.
The derivative of h(x) = ln(x) with respect to x can be found using the fact that the derivative of ln(x) is 1/x.
Now, let's apply the chain rule to find the derivative of f(x):
f'(x) = g'(h(x)) * h'(x)
Since g'(u) = 4u^3 and h'(x) = 1/x, we have:
f'(x) = 4(ln(x))^3 * 1/x
Simplifying this expression:
f'(x) = 4(ln(x))^3 / x
Therefore, the derivative of f(x) = [ln(x)]^4 is f'(x) = 4(ln(x))^3 / x.