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# 5. Quotient Rule

### Quotient Rule

Let $$f$$ and $$g$$ be differentiable functions at the point $$a.$$ If $$g(a) \neq 0,$$ the quotient rule says that $\left(\frac{f}{g}\right)'(a) = \frac{f'(a)g(a)-f(a)g'(a)}{g^2(a)}$

Proof:
By the definition of derivative, \begin{align} \left(\frac{f}{g}\right)'(a) & = \lim_{h \rightarrow 0} \frac{\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}}{h} \\ & = \lim_{h \rightarrow 0} \frac{f(a+h)g(a)-f(a)g(a+h)}{hg(a)g(a+h)} \end{align} Now add and subtract $$f(a)g(a)$$ from the numerator. \begin{align} \lim_{h \rightarrow 0} \frac{f(a+h)g(a)-f(a)g(a+h)}{hg(a)g(a+h)} & = \lim_{h \rightarrow 0} \frac{f(a+h)g(a)-f(a)g(a)+f(a)g(a)-f(a)g(a+h)}{hg(a)g(a+h)} \\ & = \lim_{h \rightarrow 0} \frac{(f(a+h)-f(a))g(a)+f(a)(g(a)-g(a+h))}{hg(a)g(a+h)} \\ & = \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \cdot \frac{g(a)}{g(a)g(a+h)} - \frac{f(a)}{g(a)g(a+h)} \cdot \frac{g(a+h)-g(a)}{h} \\ & = f'(a) \cdot \frac{g(a)}{g(a)^2} - \frac{f(a)}{g(a)^2} \cdot g'(a) \\ & = \frac{f'(a)g(a)-f(a)g'(a)}{g(a)^2} \end{align}

### Examples

Find the derivative of $$\frac{x^2}{\text{sin}(x)}.$$

Solution:
By the power rule, the derivative of $$x^2$$ is $$2x.$$ The derivative of $$\text{sin}(x)$$ is $$\text{cos}(x).$$ So, by the quotient rule, $\frac{d}{dx}\frac{x^2}{\text{sin}(x)} = \frac{2x\text{sin}(x) - x^2\text{cos}(x)}{\text{sin}(x)^2}$

Find the derivative of $$\frac{e^x}{x^3}.$$

Solution:
The derivative of $$e^x$$ is $$e^x.$$ By the power rule, the derivative of $$x^3$$ is $$3x^2.$$ So, by the quotient rule, $\frac{d}{dx}\frac{e^x}{x^3} = \frac{x^3e^x - 3x^2e^x}{x^6}$

Find the derivative of $$\frac{\text{cos}(x)}{e^x}.$$

Solution:
The derivative of $$\text{cos}(x)$$ is $$-\text{sin}(x)$$ and the derivative of $$e^x$$ is $$e^x$$. So, by the product rule, $\frac{d}{dx}\frac{\text{cos}(x)}{e^x} = \frac{-\text{sin}(x)e^x - \text{cos}(x)e^x}{e^{2x}}$ This expression can be simplified to $-\frac{\text{sin}(x) + \text{cos}(x)}{e^{x}}$