Question

let ( f(x)=(ln x)^{4} )
( f^{prime}(x)= )
( f^{prime}left(e^{3}
ight)= )
question help: video

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = u^n\) and \(u = g(x)\), then \(y^\prime=n\cdot u^{n - 1}\cdot g^\prime(x)\).
For \(f(x)=(\ln x)^4\), let \(u = \ln x\), \(n = 4\).
The derivative of \(u=\ln x\) with respect to \(x\) is \(u^\prime=\frac{1}{x}\).
Using the chain rule, \(f^\prime(x)=4(\ln x)^{4 - 1}\cdot\frac{1}{x}\).
So, \(f^\prime(x)=\frac{4(\ln x)^3}{x}\).

Step2: Evaluate \(f^\prime(e^3)\)

Substitute \(x = e^3\) into \(f^\prime(x)\).
First, find \(\ln(e^3)\). Since \(\ln(a^b)=b\ln a\) and \(\ln e = 1\), then \(\ln(e^3)=3\ln e = 3\).
Now substitute into \(f^\prime(x)\):
\(f^\prime(e^3)=\frac{4(\ln(e^3))^3}{e^3}\).
Replace \(\ln(e^3)\) with \(3\), we get \(f^\prime(e^3)=\frac{4\times3^3}{e^3}\).
Calculate \(3^3 = 27\), so \(f^\prime(e^3)=\frac{4\times27}{e^3}=\frac{108}{e^3}\).

Answer:

\(f^\prime(x)=\frac{4(\ln x)^3}{x}\), \(f^\prime(e^3)=\frac{108}{e^3}\)