Question

Differentiate the following functions from first principles x²e^x ?

Answer 1

\[\text{ Let } f\left( x \right) = x^2 e^x \]

\[ \Rightarrow f\left( x + h \right) = \left( x + h \right)^2 e^\left( x + h \right) \]

\[ = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]

\[ = \lim_{h \to 0} \frac{\left( x + h \right)^2 e^\left( x + h \right) - x^2 e^x}{h}\]

\[ = \lim_{h \to 0} \left( \frac{x^2 e^{x + h} - x^2 e^x}{h} + \frac{2xh e^\left( x + h \right)}{h} + \frac{h^2 e^\left( x + h \right)}{h} \right)\]

\[ = \lim_{h \to 0} \left( \frac{x^2 e^x \left( e^{\left( x + h \right) - x} - 1 \right)}{h} + 2x e^\left( x + h \right) + h e^\left( x + h \right) \right)\]

\[ = \lim_{h \to 0} \left( x^2 e^x \frac{\left( e^h - 1 \right)}{h} + 2x e^\left( x + h \right) + he\left( x + h \right) \right)\]

\[ = x^2 e^x + 2x e^x + 0x e^x \left[ \because \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \right]\]

\[ \therefore \frac{d}{dx}\left( x^2 e^x \right) = e^x \left( x^2 + 2x \right)\]