Question

Find dy/dx for y= 4^(lnx)

Answer 1

Answer:

$y'=\ln(4)\frac{4^{\ln(x)}}{x}$

Step-by-step explanation:

$y=4^{\ln(x)}$

Take natural log of both sides:

$\ln(y)=\ln(4^{\ln(x)})$

Use power rule of logarithms:

$\ln(y)=\ln(x)\cdot\ln(4)$

Let's differentiate.

$\frac{y'}{y}=\frac{1}{x} \cdot \ln(4)$

I applied chain rule on the left side and I apply constant multiple rule to the right side.

Let's multiply both sides by $y$:

$y'=y \cdot \frac{1}{x} \cdot \ln(4)$

We started with $y=4^{\ln(x)}$ so let's make that replacement:

$y'=4^{\ln(x)} \cdot \frac{1}{x} \cdot \ln(4)$

Let's simplify it a bit:

$y'=\ln(4)\frac{4^{\ln(x)}}{x}$

Answer 2

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