Solving an ln derivative

[grapz]

y = ( 4 ^ x ) (ln x )

is this right?

ln y = ln 4 x ( ln x)
1/y y ' = ln 4 ln x + 1/x( x ln 4)
y' = [ln 4 ln x + 1/x(x ln 4) ] ( 4^x )(lnx)

 

[morson]

..What? Use the product rule.

\(\displaystyle \L\ \frac{dy}{dx}\ = log_e (4) 4^x log_e (x) + \frac{4^x}{x}\\)

 

[stapel]

Solving an ln derivative
y = ( 4 ^ x ) (ln x )

I'm sorry, but what do you mean by "solving" a derivative? Are you trying to find the max/min points (by setting the derivative equal to zero and solving), or something else? What were the exact instructions?

Thank you!

Eliz.

 

[grapz]

sorry i meant to solve for the derviative of that equation, meaning find dy/dx

thanks

and i did use product rule but , i dont't hink i know how to find the derivative of 4^x , that is why i had to do ln of both sides

 

[pka]

\(\displaystyle \L \begin{array}{l}
a > 0 \\
g(x) = a^x \quad \Rightarrow \quad g'(x) = a^x \ln (a) \\
\end{array}\)