Question

if ( f(x)=7 x^{ln (x)} ), find ( f^{prime}(8) ). ( f^{prime}(8)= )

Explanation:

Step1: Take natural logarithm on both sides

Let \( y = 7x^{\ln(x)} \), then \( \ln y=\ln(7x^{\ln(x)})=\ln7+\ln(x^{\ln(x)})=\ln7 + (\ln x)^2 \).

Step2: Differentiate both sides with respect to \( x \)

Using the chain - rule, \(\frac{1}{y}y' = 2\frac{\ln x}{x}\).

Step3: Solve for \( y' \)

Multiply both sides by \( y \), so \( y'=y\times2\frac{\ln x}{x}\). Since \( y = 7x^{\ln(x)} \), then \( y'=7x^{\ln(x)}\times2\frac{\ln x}{x}=14x^{\ln(x)- 1}\ln x \).

Step4: Substitute \( x = 8 \)

When \( x = 8 \), \( f'(8)=14\times8^{\ln(8)-1}\ln8 \).
First, \( \ln8=\ln(2^{3}) = 3\ln2\approx3\times0.693 = 2.079 \), and \( 8^{\ln(8)-1}=8^{\ln8}\times8^{-1}=\frac{8^{\ln8}}{8}\).
Since \( a^{\ln b}=b^{\ln a} \), then \( 8^{\ln8}=8^{\ln8}=e^{\ln8\times\ln8}\), but another way: \( y = x^{\ln x}\), when \( x = 8 \), \( y=8^{\ln8}\). Also, using \( y' \) formula \( y'=x^{\ln x}\times2\frac{\ln x}{x}\) (from \( y = x^{\ln x}\), \( \ln y=(\ln x)^2\), \( \frac{y'}{y}=2\frac{\ln x}{x}\)).

Another approach:
Let \( u=\ln x \), \( y = 7e^{u\cdot u}=7e^{u^{2}}\), \( \frac{dy}{du}=14ue^{u^{2}}\), \( \frac{du}{dx}=\frac{1}{x}\).
By the chain - rule \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=14\frac{\ln x}{x}e^{(\ln x)^{2}}=14x^{\ln x - 1}\ln x \).
When \( x = 8 \):
\( f'(8)=14\times8^{\ln8 - 1}\ln8=14\times\frac{8^{\ln8}}{8}\ln8\).
Since \( 8^{\ln8}=e^{\ln8\times\ln8}\approx e^{(2.079)^{2}}\approx e^{4.322}\approx75.27\)
\( f'(8)=14\times\frac{75.27}{8}\times2.079\approx14\times9.409\times2.079\approx14\times19.56\approx273.84\)

Answer:

\(273.84\)