Calculating the Derivative of an Exponential Function with Logarithms?

[karush]

Find $f'(x)$
$\displaystyle
f(x)={3}^{2x+5}+\log_3(x^2+4)$

Didn't know how to do the $\log_3(x^2+4)$

 

[Greg]

$$y=\log_3(x^2+4)$$

$$3^y=x^2+4$$

$$y\ln(3)=\ln(x^2+4)$$

$$y'\ln(3)=\frac{2x}{x^2+4}$$

$$y'=\frac{2x}{\ln(3)\left(x^2+4\right)}$$

 

[karush]

I'm in math heaven

 

[Greg]

The change of base formula for logs* may also be used:

$$\begin{align*}y&=\log_3(x^2+4) \\
&=\frac{\ln(x^2+4)}{\ln(3)} \\
y'&=\frac{2x}{\ln(3)(x^2+4)}\end{align*}$$

*Proof of the change of base formula:

$$\begin{align*}y&=\log_a(b) \\
a^y&=b \\
y\log_c(a)&=\log_c(b) \\
&y=\frac{\log_c(b)}{\log_c(a)}\end{align*}$$

 

[HallsofIvy]

Or just use the fact that \frac{d(log_a(x))}{dx}= \frac{1}{log_e(a)}\frac{1}{x}.

That follows from the "change of base" formula greg1313 gave: \frac{d(log_a(x))}{dx}= \frac{d\left(\frac{log_e(x)}{log_e(a)}\right)}{dx}= \frac{1}{log_e(a)}\frac{dlog_e(x)}{dx}= \frac{1}{log_e(a)}\frac{1}{x}.

Generally, \frac{da^x}{dx}= ln(a) a^x and \frac{dlog_a(x)}{dx}= \frac{1}{ln(a)}\frac{1}{x}.

 

[karush]

Thanks everyone. Major help