Question

Find f ′(x) for f(x) = e^(x)ln(x).

Answer 1

Three derivative rules to have in mind:
\[h(x)=f(x).g(x) \iff h'(x)=f'(x)g(x)+f(x)g'(x)\]
\[f(x)=e^x \iff f'(x)=e^x\]
\[f(x)=\ln(x) \iff f'(x)=\frac{ 1 }{ x }\]

Answer 2

ok that's not even answer choice though so im confused

Answer 3

It isn't the answer, these are the derivation rules you can use to finde the derivative of the given function.

Answer 4

ohh ok

Answer 5

This implies, that having the function \(f(x)=e^x lnx \) we can view it as a product of two functions, one being \(e^x\) and the other \(\ln x\) and as I wrote above we can use the product rule for derivation to derive the function \(f(x)=e^x lnx \) because we now recognized it is a product of two functions:
\[f(x)= e^xln x\]
the product rule for derivation can be memorized by thinking:
"the derivative of the first function multiplied the second without deriving, plus, the first function without deriving multiplied the derivative of the second function".
\[f'(x)=(e^x)(lnx)+(e^x)(\frac{ 1 }{ x })\]
Notice that I wrote the derivatives for \(e^x\) and for \(\ln x\) above, so those derivatives shouldn't represent a problem.
Now, as a task I leave to you is to simplify \(f'(x)=(e^x)(lnx)+(e^x)(\frac{ 1 }{ x })\) in order to obtain a simpler answer.

Answer 6

Use \ln for a nicer natural logarithm typesetting.
\[
ln(x)\quad\text{Using "ln(x)"}\\
\ln(x)\quad\text{Using "\ln(x)"}
\]

Answer 7

Same with sine, cosine and tangent. Use \sin, \cos and \tan instead of sin, cos and tan.