Question

Differentiate the following functions from first principles:

x²e^x

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(x\right)=x^2{e}^x \\ \Rightarrow f\left(x+h\right)={\left(x+h\right)}^2{e}^{\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2{e}^{\left(x+h\right)}-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\left(\frac{x^2{e}^{x+h}-x^2{e}^x}{h}+\frac{2xh{e}^{\left(x+h\right)}}{h}+\frac{h^2{e}^{\left(x+h\right)}}{h}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{x^2{e}^x\left(e^{\left(x+h\right)-x}-1\right)}{h}+2x{e}^{\left(x+h\right)}+h{e}^{\left(x+h\right)}\right) \\ =\underset{h\to 0}{\lim }\left(x^2{e}^x\frac{\left(e^h-1\right)}{h}+2x{e}^{\left(x+h\right)}+he\left(x+h\right)\right) \\ =x^2{e}^x+2x{e}^x+0x{e}^x\left[\because \underset{x\to 0}{\lim }\frac{e^x-1}{x}=1\right] \\ \therefore \frac{d}{dx}\left(x^2{e}^x\right)=e^x\left(x^2+2x\right) \end{gathered}$$