Question

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

Answer 1

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

Explanation:

`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.`

Spread the Joy Maths.!

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!)`

A general Maclaurin series is given by:

`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!)`.

Equating their coefficients:

`f^6(0)/(6!)x^6=x^6/(5!)`

`f^6(0)=(6!)/(5!)=6`