How do you differentiate #g(x) = e^(2x)ln3x# using the product rule?

Question

Emilia Aguilar · Accepted Answer

I found: #e^(2x)[2ln(3x)+1/x]#

According to the Product Rule, if you have:

#g(x)=f(x)*h(x)# then #g'(x)=f'(x)h(x)+f(x)h'(x)#

In your situation, after deriving each function, you must additionally apply the Chain Rule (highlighted in red) to obtain:

#g'(x)=color(red)(2e^(2x))ln(3x)+e^(2x)color(red)(1/(3x)*3)=#

#=e^(2x)[2ln(3x)+1/x]#

Emily Ammerman · Answer

undefined To differentiate $ g(x) = e^{2x} \ln(3x) $ using the product rule, follow these steps:

1. Identify the two functions being multiplied: $ f(x) = e^{2x} $ and $ h(x) = \ln(3x) $.
2. Apply the product rule: $ g'(x) = f'(x)h(x) + f(x)h'(x) $.
3. Find the derivatives of each function: $ f'(x) = 2e^{2x} $ and $ h'(x) = \frac{1}{x} $.
4. Substitute these derivatives into the product rule formula: $ g'(x) = 2e^{2x} \ln(3x) + e^{2x} \cdot \frac{1}{x} $.
5. Simplify the expression if necessary.