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)