Answer to Question #315587 in Calculus for Aune

Question

If 𝑓(𝑥) is a differentiable and 𝑔(𝑥) = 𝑥 𝑓(𝑥) use the definition of the derivative to show

that 𝑔′(𝑥) = 𝑥𝑓'(𝑥) + 𝑓(𝑥).

Answer 1

Given a function $f(x)$ , the derivative (by definition) is:

$$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

Now , given $𝑔(𝑥) = 𝑥 𝑓(𝑥)$ ,

$$\begin{aligned} g'(x) = & \lim_{h \rightarrow 0} \frac{(x+h)f(x+h) - xf(x)}{h}\\ =& \lim_{h \rightarrow 0} \frac{xf(x+h)+hf(x+h) - xf(x)}{h}\\ =& \lim_{h \rightarrow 0} \frac{xf(x+h)- xf(x)+hf(x+h)}{h}\\ =& \lim_{h \rightarrow 0} \frac{x(f(x+h)- f(x))+hf(x+h)}{h}\\ =& \lim_{h \rightarrow 0} \frac{x(f(x+h)- f(x))}{h} + \lim_{h \rightarrow 0} \frac{hf(x+h)}{h}\\ =& \lim_{h \rightarrow 0} x\Bigg(\frac{f(x+h)- f(x)}{h}\Bigg) + \lim_{ h\rightarrow 0} f(x+h)\\ g'(x)= & xf'(x) + f(x) \end{aligned}$$

Q.E.D