Proof of quotient rule $(\frac{f}{g})'(x_{0})=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}$

Question

$$\left(\frac{f}{g}\right)'(x_{0})=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}$$

So, $\frac{1}{g}.f=\frac{f}{g}$, then $$\frac{f}{g}'(x_0)=\frac{f(x)\frac{1}{g(x)}-f(x_0)\frac{1}{g(x_0)}}{x-x_0}=f(x)\frac{\frac{1}{g(x)}-\frac{1}{g(x_0)}}{x-x_0}-\frac{1}{g(x_0)}\frac{f(x)-f(x_0)}{x-x_0}$$ I took the limit of everything as $x \to x_0$ $$=f(x_0)\frac{1}{g'(x_0)}-\frac{1}{g(x_0)}f'(x_0)=\frac{f(x_0)}{g'(x_0)}-\frac{f'(x_0)}{g(x_0)}=\frac{g(x_0)f(x_0)-f'(x_0)g'(x_0)}{g'(x_0)g(x_0)}$$ which clearly it's fake.

Where did I go wrong? Thank you!

Answer 1

You are close. Note that instead of cancelling the term $\frac{\frac{f(x)}{g(x_0)}}{x-x_0}$ you actually subtract it twice. Second, note that the $$\lim_{x \rightarrow x_0} \frac{\frac{1}{g(x)} - \frac{1}{g(x_0)}}{x-x_0} = \left(\frac{1}{g(x_0)}\right)'\neq \frac{1}{g'(x_0)}.$$

Instead of doing it this way, I like to prove the quotient rule using the product rule: $\frac{f}{g} = f \cdot g^{-1}$. $\frac{d}{dx} (fg^{-1}) = f'g^{-1} - fg'g^{-2} = \frac{f'g - fg'}{g^2}.$

Answer 2

$(1/g)'\neq 1/g'$ this is where you made the mistake.

Answer 3

There is another way to do it : logarithmic differentiation $$y=\frac{f(x)}{g(x)}\implies \log(y)=\log(f(x))-\log(g(x))$$ Differentiate both sides $$\frac{y'}y=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}=\frac{f'(x)\,g(x)-f(x)\,g'(x) } {f(x)\,g(x) }$$ $$y'=y \,\frac{y'}y=\frac{f(x)}{g(x)}\times \frac{f'(x)\,g(x)-f(x)\,g'(x) } {f(x)\,g(x) }$$ Simplify.