Answer:
[tex]y'=\ln(4)\frac{4^{\ln(x)}}{x}[/tex]
Step-by-step explanation:
[tex]y=4^{\ln(x)}[/tex]
Take natural log of both sides:
[tex]\ln(y)=\ln(4^{\ln(x)})[/tex]
Use power rule of logarithms:
[tex]\ln(y)=\ln(x)\cdot\ln(4)[/tex]
Let's differentiate.
[tex]\frac{y'}{y}=\frac{1}{x} \cdot \ln(4)[/tex]
I applied chain rule on the left side and I apply constant multiple rule to the right side.
Let's multiply both sides by [tex]y[/tex]:
[tex]y'=y \cdot \frac{1}{x} \cdot \ln(4)[/tex]
We started with [tex]y=4^{\ln(x)}[/tex] so let's make that replacement:
[tex]y'=4^{\ln(x)} \cdot \frac{1}{x} \cdot \ln(4)[/tex]
Let's simplify it a bit:
[tex]y'=\ln(4)\frac{4^{\ln(x)}}{x}[/tex]
Answer:
dy/dx = (4^lnx)(ln4)/x
Step-by-step explanation:
lny = (lnx)ln4
(1/y)y' = (ln4)/x
y' = y(ln4)/x
y' = (4^lnx)(ln4)/x
