How do I find the derivative of #ln(e^(4x)+3x)#?

Question

Isabella Ackerman · Accepted Answer

#(f(g(x)))'=(4e^(4x)+3)/(e^(4x)+3x)#

We can find the derivative of this function using chain rule that says:

#color(blue)((f(g(x)))'=f'(g(x))*g'(x))#

Let us decompose the given function into two functions #f(x)# and #g(x)# and find their derivatives as follows:

#g(x)=e^(4x)+3x# #f(x)=ln(x)#

Let's find the derivative of #g(x)# Knowing the derivative of exponential that says: #(e^(u(x)))'=(u(x))'*e^(u(x))# So, #(e^(4x))'=(4x)'*e^(4x)=4e^(4x)# Then , #color(blue)(g'(x)=4e^(4x)+3)#

Now Lets find #f'(x)#

#f'(x)=1/x# According to the property above we have to find #f'(g(x))# so let's substitute #x# by #g(x)# in #f'(x)# we have:

#f'(g(x))=1/g(x)# #color(blue)(f'(g(x))=1/(e^(4x)+3x))# Therefore, #(f(g(x)))'=(1/(e^(4x)+3x))*(4e^(4x)+3)#

#color(blue)((f(g(x)))'=(4e^(4x)+3)/(e^(4x)+3x))#

Cameron Alexander · Answer

undefined To find the derivative of $ \ln(e^{4x} + 3x) $, you can use the chain rule and the fact that the derivative of $ \ln(u) $ with respect to $ u $ is $ \frac{1}{u} $.

Let $ u = e^{4x} + 3x $.

Then, using the chain rule, the derivative of $ \ln(u) $ with respect to $ x $ is:

$$ \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} $$

First, find $ \frac{du}{dx} $:

$$ \frac{du}{dx} = \frac{d}{dx} (e^{4x} + 3x) $$
$$ = \frac{d}{dx} e^{4x} + \frac{d}{dx} 3x $$
$$ = 4e^{4x} + 3 $$

Now, substitute $ \frac{du}{dx} = 4e^{4x} + 3 $ into the derivative formula:

$$ \frac{d}{dx} \ln(u) = \frac{1}{e^{4x} + 3x} \cdot (4e^{4x} + 3) $$

Therefore, the derivative of $ \ln(e^{4x} + 3x) $ with respect to $ x $ is:

$$ \frac{4e^{4x} + 3}{e^{4x} + 3x} $$