Find the nth Derivatives of lnx, xe^x, and e^x cosx

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Hockeyman

Junior Member
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Dec 8, 2005
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The instructions say to find the nth derivative of each of the following functions:

A. lnx

For this one I used the natural log rule and made it 1/x, so my equation was y=1/x

B. xe^x

I'm not sure how to start this one

C. e^x cosx

Not sure how to start this one either, could someone tell me if the first one is right and how to start the other two?
 
I will give you some help on #1.
\(\displaystyle \L \[
D_x ^k \left[ {\ln (x)} \right] = \frac{{( - 1)^{k - 1} \left( {k - 1} \right)!}}{{x^k }}\)
 
If you take the derivatives you can see the pattern.

ln(x)

\(\displaystyle \L\\f'(x)=1/x\)

\(\displaystyle \L\\f''(x)=\frac{-1}{x^{2}}\)

\(\displaystyle \L\\f'''(x)=\frac{2}{x^{3}}\)

\(\displaystyle \L\\f^{4}(x)=\frac{-6}{x^{4}}\)

\(\displaystyle \L\\f^{5}(x)=\frac{24}{x^{5}}\)

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Do the same for \(\displaystyle \L\\xe^{x}\). It's an easy one.
 
I get how you find the denominator in the first problem. The exponent matches the term number. How do you find the numerator though?

For the second one i got the following:

1. e^x + xe^x

2. e^x +(e^x + xe^x)

3. e^x + {e^x +(e^x + xe^x)

so what would the rule be?
 
Now factor and see the pattern

\(\displaystyle \L\\xe^{x}\)

\(\displaystyle \L\\f'(x)=(x+1)e^{x}\)

\(\displaystyle \L\\f''(x)=(x+2)e^{x}\)

\(\displaystyle \L\\f'''(x)=(x+3)e^{x}\)

\(\displaystyle \L\\f^{4}(x)=(x+4)e^{x}\)

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ok i get that one, now for the first one i have
-(k-1) x (numerator of the previous term) / x^k

How do you right that mathmatically?
 
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