Let $\mathrm{f}(x)=x^3\mathrm{e}^{-x}$. Using the definition of a derivative, show from first prin

Question

Let $\mathrm{f}(x)=x^3\mathrm{e}^{-x}$. Using the definition of a derivative, show from first principles that the derivative of $x^3$ is $3x^2$. Prove by induction that, for every positive integer $n$, $$\begin{aligned}\mathrm{f}^{(n)}(x)=(-1)^n\mathrm{e}^{-x}[&x^3-3nx^2+3n(n-1)x\&-n(n-1)(n-2)].\end{aligned}$$

Accepted Answer

(a) $\dfrac{\mathrm{d}}{\mathrm{d}x}(x^3)=3x^2$ (b) $\mathrm{f}^{(n)}(x)=(-1)^n\mathrm{e}^{-x}[x^3-3nx^2+3n(n-1)x-n(n-1)(n-2)]$, $n\ge1$

Cubic decaying exponential derivative pattern

Cubic decaying exponential derivative pattern

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