Question

Let f(x) = x^9 e^x then the ninth derivative of f(x) at x = 0 is given by?

(a) 9!

(b) 9! * e^9

(c) 10!

(d) 21!

I had been asked this question by my college director while I was bunking the class.

I would like to ask this question from The nth Derivative of Some Elementary Functions topic in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) 9!

To explain: The key here is to expand e^x as a mclaurin series and then multiply it by x^9

We have e^x=\(\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}…\infty\)

Now multiplying it by x^9 we get

f(x)=\(\frac{x^9}{0!}+\frac{x^{10}}{1!}+\frac{x^{11}}{2!}…\infty\)

This simplifies into polynomial differentiation and the ninth derivative is as follows

 f^(9)(x)=\(9!+\frac{10!\times x}{1!}+\frac{p_9^{11}\times x^2}{2!}…\infty\)

Substituting x = 0 gives us the result

 f^(9)(0) = 9!.