> [!summary] This proof uses the definition of derivatives and limits rules to prove the quotient rule. > Key Result: $\frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$ >[!info]+ Read Time **⏱ 2 mins** # Mathematical Proof > [!warning] Assumptions To prove the derivative quotient rule using [[Direct Proof|direct proof]] assume the following: > - The definition of a [[Derivative|derivative]] is $\frac{dy}{dx} = f'(x) = \displaystyle \lim_{ h \to 0 } \frac{f(x+h)-f(x)}{h}$ > - The [[Analytical Limits Rules|limit sum rule]] is $\displaystyle \lim_{ x \to a } [f (x)+g (x)]=L+M$ > - The [[Greatest Common Divisor (gcd)|greatest common denominator]] of $\frac{\frac{f (x+h)}{g (x+h)}-\frac{f (x)}{g (x)}}{h}=\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)h}$ Prove $\frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$ $ \begin{array}{c} \\ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \displaystyle \lim_{ h \to 0 } \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h} \\ = \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)h} \\ \\ \text{Using the limit sum rule spilt the limit into two parts:} \\ \displaystyle \lim_{ h \to 0 } \frac{1}{g(x+h)g(x)}\cdot \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x)-f(x)g(x+h)}{h} \\ \\ \text{Simplify the second limit by adding and subtracting by $f(x)g(x)$} \\ \begin{align*} &= \displaystyle \lim_{ h \to 0 } \frac{f(x+h)g(x)-f(x)g(x+h)+ f(x)g(x)-f(x)g(x)}{h} \\ &= \displaystyle \lim_{ h \to 0 } \frac{g(x)f(x+h)+g(x)f(x)}{h} - \displaystyle \lim_{ h \to 0 } \frac{f(x)g(x+h)+f(x)g(x)}{h} \\ &= g(x) \displaystyle \lim_{ h \to 0 } \frac{f(x+h)+f(x)}{h} - f(x)\displaystyle \lim_{ h \to 0 } \frac{g(x+h)+g(x)}{h} \end{align*} \\ \\ \text{So combining the first and second term:} \\ \displaystyle \lim_{ h \to 0 } \frac{1}{g(x+h)g(x)} \cdot \left[g(x) \displaystyle \lim_{ h \to 0 } \frac{f(x+h)+f(x)}{h} - f(x)\displaystyle \lim_{ h \to 0 } \frac{g(x+h)+g(x)}{h} \right] \\ \\ \text{By definition this becomes} \\ \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} \end{array} $ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/jxxzbMxihjQ?si=KzNQOWNiejPb7RzM" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> --- <!-- Light Mode Newsletter Embed --> <div class="mm-form-light"> <iframe src="https://updates.cyberleadhub.com/widget/form/Y0kpQVpjJQuxEfX59m17" id="inline-Y0kpQVpjJQuxEfX59m17" title="Join Math & Matter Newsletter (Light)" data-height="900" scrolling="no" allowtransparency="true" loading="lazy" style="width:100%;height:350px;border:none;border-radius:10px;background:transparent;overflow:hidden" ></iframe> </div> <!-- Dark Mode Newsletter Embed --> <div class="mm-form-dark"> <iframe src="https://updates.cyberleadhub.com/widget/form/lbeDLm24VjuaFxhjccA1" id="inline-lbeDLm24VjuaFxhjccA1" title="Join Math & Matter Newsletter (Dark)" data-height="900" scrolling="no" allowtransparency="true" loading="lazy" style="width:100%;height:350px;border:none;border-radius:10px;background:transparent;overflow:hidden" ></iframe> </div> <!-- Provider script (only once) --> <script src="https://updates.cyberleadhub.com/js/form_embed.js"></script>