Question #315587

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).

1
Expert's answer
2022-03-28T07:05:21-0400

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


f(x)=limh0f(x+h)f(x)hf^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

Now , given 𝑔(𝑥)=𝑥𝑓(𝑥)𝑔(𝑥) = 𝑥 𝑓(𝑥) ,


g(x)=limh0(x+h)f(x+h)xf(x)h=limh0xf(x+h)+hf(x+h)xf(x)h=limh0xf(x+h)xf(x)+hf(x+h)h=limh0x(f(x+h)f(x))+hf(x+h)h=limh0x(f(x+h)f(x))h+limh0hf(x+h)h=limh0x(f(x+h)f(x)h)+limh0f(x+h)g(x)=xf(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

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