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# 4. Product Rule

### Product Rule

Let $$f$$ and $$g$$ be differentiable functions at the point $$a.$$ The product rule says that $(fg)'(a) = f'(a)g(a) + f(a)g'(a)$

Proof:
By the definition of derivative, $(fg)'(a) = \lim_{h \rightarrow 0} \frac{f(a+h)g(a+h) - f(a)g(a)}{h}$ Add and subtract $$f(a)g(a+h),$$ then factor. \begin{align} \lim_{h \rightarrow 0} \frac{f(a+h)g(a+h) - f(a)g(a)}{h} & = \lim_{h \rightarrow 0} \frac{f(a+h)g(a+h) - f(a)g(a+h) + f(a)g(a+h) - f(a)g(a)}{h} \\ & = \lim_{h \rightarrow 0} \frac{(f(a+h) - f(a))g(a+h) + f(a)(g(a+h) - g(a))}{h} \\ & = \lim_{h \rightarrow 0} \frac{(f(a+h) - f(a))g(a+h)}{h} + \lim_{h \rightarrow 0} \frac{f(a)(g(a+h) - g(a))}{h} \end{align} Now use $$\lim_{h \rightarrow 0} g(a+h) = g(a)$$ and the definition of the derivatives of $$f$$ and $$g$$ to get $\lim_{h \rightarrow 0} \frac{(f(a+h) - f(a))g(a+h)}{h} + \lim_{h \rightarrow 0} \frac{f(a)(g(a+h) - g(a))}{h} = f'(a)g(a)+f(a)g'(a)$

### Examples

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

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 product rule, $\frac{d}{dx}x^2\text{sin}(x) = 2x\text{sin}(x) + x^2\text{cos}(x)$

Find the derivative of $$x^3e^x.$$

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

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

Solution:
The derivative of $$e^x$$ is $$e^x$$ and the derivative of $$\text{sin}(x)$$ is $$\text{cos}(x).$$ So, by the product rule, $\frac{d}{dx}e^x\text{sin}(x) = e^x\text{sin}(x) + e^x\text{cos}(x)$