How do you find #f^6(0)# where #f(x)=xe^x#?

Answer 1

#f^((6))(0)=(0+6)e^0=6.#

#f(x)=xe^x#
By the Product Rule, #f'(x)=x(e^x)'+(e^x)(x)'#
#:. f'(x)=xe^x+e^x=(x+1)e^x..........................(1)#
This means that, #{xe^x}'=(x+1)e^x.............(star)#
#"Now, "f''(x)={f'(x)}'={xe^x+e^x}'={xe^x}'+(e^x)'#
#=(x+1)e^x+e^x,....................[because, (star)]#
#:. f''(x)=(x+2)e^x...................................(2)#
#"As, "f''(x)=xe^x+2e^x," we have, by "(star),#
#f'''(x)={f''(x)}'={xe^x}'+(2e^x)'=(x+1)e^x+2e^x, i.e.,#
#f'''(x)=(x+3)e^x#.......................................(3)#
#"Generalising, "f^((n))(x)=(x+n)e^x, n in NN.#
#"In Particular, "f^((6))(x)=(x+6)e^x," giving,"#
#f^((6))(0)=(0+6)e^0=6.#

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Answer 2
We can also use the Maclaurin series for #e^x#:
#e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...#
Multiplying this by #x# we see that:
#xe^x=x(1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...)=x+x^2+x^3/(2!)+x^4/(3!)+x^5/(4!)+...#

Or:

#xe^x=sum_(n=0)^oox^(n+1)/(n!)#

The following yields a general Maclaurin series:

#f(x)=sum_(n=0)^oof^n(0)/(n!)x^n#
So when #n=6# the #6#th term of any general Maclaurin series is #f^6(0)/(6!)x^6#.
Also note that the #n=5# term of the #f(x)=xe^x# series is #x^6/(5!)#.

calculating the coefficients of each:

#f^6(0)/(6!)x^6=x^6/(5!)#
#f^6(0)=(6!)/(5!)=6#
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Answer 3

To find ( f^6(0) ) where ( f(x) = xe^x ), you would:

  1. Calculate the first derivative of ( f(x) ) with respect to ( x ).
  2. Then, calculate the second derivative of ( f(x) ).
  3. Repeat this process until you've calculated the sixth derivative of ( f(x) ).
  4. Evaluate the sixth derivative of ( f(x) ) at ( x = 0 ) to find ( f^6(0) ).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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