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- 3.2 Quotient Ruleap Calculus Equation
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3.2 Quotient Ruleap Calculus Equation
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3.2 Quotient Ruleap Calculus Solver
3.2 The Product Rule and the Quotient Rule Math 1271, TA: Amy DeCelles 1. Overview You need to memorize the product rule and the quotient rule. And, more than that actually: you need to internalize them. The best way to do that is just by practicing until you can use them without even thinking about it. Product Rule: (fg)0= f0g + fg0. 3 - Rules of Differentiation Notes 3.2 Key. Powered by Create your own unique website with customizable templates. Quotient rule the derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function: sum rule.
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3.2 Quotient Ruleap Calculus Calculator
(a) $u(x)=f(x)g(x)$. Therefore, $$u'(x)=f'(x)g(x)+f(x)g'(x)$$ That means, $$u'(1)=f'(1)g(1)+f(1)g'(1)$$ From the graph, we can easily see that $f(1)=2$ and $g(1)=1$ To find $f'(1)$, we need to find the slope of the tangent line to $f$ at $x=1$ However, from $x=0$ to $x=2$, graph $f$ is a straight line. So, the slope of the tangent line to $f$ at $x=1$ is $$f'(1)=frac{f(2)-f(0)}{2-0}=frac{4-0}{2-0}=2$$ Similarly, $g'(1)$ is also the slope of tangent line to $g$ at $x=1$. Also, from $x=0$ to $x=2$, graph $g$ is also a straight line. So, $$g'(1)=frac{g(2)-g(0)}{2-0}=frac{0-2}{2-0}=-1$$ Therefore, $$u'(1)=2times1+2times(-1)=0$$ (b) $v(x)=f(x)/g(x)$. Therefore, $$v'(x)=frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ That means, $$v'(5)=frac{f'(5)g(5)-f(5)g'(5)}{[g(5)]^2}$$ From the graph, we can easily see that $f(5)=3$ and $g(5)=2$ To find $f'(5)$, we need to find the slope of the tangent line to $f$ at $x=5$ However, from $x=2$ to $x=5$, graph $f$ is a straight line. So, the slope of the tangent line to $f$ at $x=5$ is $$f'(5)=frac{f(5)-f(2)}{5-2}=frac{3-4}{3}=frac{-1}{3}$$ Similarly, $g'(5)$ is also the slope of tangent line to $g$ at $x=5$. Also, from $x=2$ to $x=5$, graph $g$ is also a straight line. So, $$g'(5)=frac{g(5)-g(2)}{5-2}=frac{2-0}{3}=frac{2}{3}$$ Therefore, $$v'(5)=frac{-1/3times2-3times(2/3)}{2^2}=frac{-2/3-2}{4}=frac{-2}{3}$$