Find the derivative of
y=(4x-1)(e^x)(ln(3x+1))

Question

Find the derivative of 
y=(4x-1)(e^x)(ln(3x+1))

anonymous · Answer

do you know how to use the chain rule?

lin.ivory · Answer

i do! ive worked on it for 30 mins with product rule but my answer is different from the key

anonymous · Answer

could you show your work for me, or do you want me to do it and show my work for you?

lin.ivory · Answer

can you show your work please? heres what i worked out from before thats apparently incorrect

$$y'=(4xe ^{x}+4e ^{x}-e ^{x})(\ln(3x+1))$$
$$=(4xe ^{x}+4e ^{x}+4e ^{x}-e ^{x})\ln(3x+1)+(4e ^{x}+4x-1)e ^{x})(\frac{ 3 }{ 3x+1 })$$
$$=(4xe ^{x}+7e ^{x})\ln(3x+1)+e ^{x}(4+4x-1)(\frac{ 3 }{ 3x+1 })$$
$$=e ^{x}(4x+7)\ln(3x+1)+e ^{x}(4x+3)(\frac{ 3 }{ 3x+1 })$$

anonymous · Answer

try this: y(x)=f(x)g(x)h(x)
product rule with three terms
and you get: y'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)

anonymous · Answer

sorry for the delay

lin.ivory · Answer

thats alright!!! THANK YOU SO MUCH!!! :D :D

anonymous · Answer

you're welcome

anonymous · Answer

anything else?