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Mathematical Expression Editor
Here we compute derivatives of products and quotients of functions
1 The product rule
Consider the product of two simple functions, say
\[ f(x)\cdot g(x) \]
where \(f(x)=x^2+1\) and \(g(x)=x^3-3x\). An obvious guess for
the derivative of \(f(x)g(x)\) is the product of the derivatives: \begin{align*} f'(x)g'(x) &= (2x)(3x^2-3)\\ &= 6x^3-6x. \end{align*}
Is this guess correct? We can check by rewriting \(f\) and \(g\) and doing the calculation in a
way that is known to work. Write with me \begin{align*} f(x)g(x) &= (x^2+1)(x^3-3x)\\ &=x^5-3x^3+x^3-3x\\ &=x^5-2x^3-3x. \end{align*}
Let’s return to the example with which we started.
Let \(f(x)=(x^2+1)\) and \(g(x)=(x^3-3x)\). Compute:
\[ \frac {d}{dx} f(x)g(x) \]
Write with me \begin{align*} \frac {d}{dx} f(x)g(x) &= f(x)g'(x) + f'(x)g(x)\\ &=(x^2+1)(\answer [given]{3x^2-3}) + (\answer [given]{2x})(x^3-3x). \end{align*}
We could stop here, but we should show that expanding this out recovers our
previous result. Write with me \begin{align*} (x^2+1)&(3x^2-3) + 2x(x^3-3x)\\ &= 3x^4-3x^2 +3x^2 -3 + 2x^4-6x^2\\ &=\answer [given]{5x^4-6x^2-3}, \end{align*}
which is precisely what we obtained before.
Now that we are pros, let’s try one more example.
Compute:
\[ \frac {d}{dx}(xe^x-e^x) \]
Using the sum rule and the product rule, write with me \begin{align*} \frac {d}{dx} \left (xe^x-e^x \right ) &=\frac {d}{dx} \left (xe^x\right )-\frac {d}{dx} e^x\\ &=(xe^x+e^x)-e^x\\ &=\answer [given]{x e^x}. \end{align*}
2 The quotient rule
We’d like to have a formula to compute
\[ \frac {d}{dx} \frac {f(x)}{g(x)} \]
This brings us to our next derivative
rule.
The quotient rule If \(f\) and \(g\) are differentiable, then
Write with me \begin{align*} \frac {d}{dx} \frac {x^2+1}{x^3-3x} &= \frac {2x(\answer [given]{x^3-3x})-(\answer [given]{x^2+1})(3x^2-3)}{(\answer [given]{x^3-3x})^2}\\ &=\frac {-x^4-6x^2+3}{(\answer [given]{x^3-3x})^2}. \end{align*}
Compute:
\[ \frac {d}{dx} \frac {\sin {x}}{x} \]
Write with me \begin{align*} \frac {d}{dx} \frac {\sin {x}}{x} &= \frac {\cos {x}\cdot x-\sin {x}\cdot 1}{x^2} \end{align*}
It is often possible to calculate derivatives in more than one way, as we have already
seen. Since every quotient can be written as a product, it is always possible
to use the product rule to compute the derivative, though it is not always
simpler.
Compute:
\[ \frac {d}{dx} \frac {625-x^2}{\sqrt {x}} \]
in two ways. First using the quotient rule and then using the product rule.
First, we’ll compute the derivative using the quotient rule. Write with me
Suppose we have two functions, \(f\),
and \(g\), and we know that \(f(4) = 3\), \(f'(4) = 5\), \(g(4) = -2\), and \(g'(4) = 2\). What is the slope of the tangent line to the curve \(y=\frac {f(x)}{g(x)}\)
at the point where \(x = 4\)?
The slope of the tangent line to the curve \(y=\frac {f(x)}{g(x)}\) at \(x = 4\) is given by
\( \bigg [ \frac {d}{dx}\frac {f(x)}{g(x)} \bigg ]_{x=4}\).
By the QuotientRule, this derivative is given by \begin{align*} \bigg [ \frac {d}{dx}\frac {f(x)}{g(x)} \bigg ]_{x=4} &=\bigg [ \frac {f'(x)g(x)-f(x)g'(x)}{\left (g(x)\right )^2} \bigg ]_{x=4}\\ &=\frac {f'(4)g(4)-f(4)g'(4)}{\left (g(4)\right )^2}\\ &=\frac {5(-2)-3\cdot 2}{\left (-2\right )^2}\\ &=\frac {-10-6}{4}\\ &=4 \end{align*}
Suppose we have two functions, \(f\), and \(g\), and we know that \(f(4) = 3\), \(f'(4) = 5\), \(g(4) = -2\), and \(g'(4) = 2\). What is the
slope of the tangent line to the curve \(y=\frac {xf(x)}{g(x)}\) at the point where \(x = 4\)?
The slope of the tangent
line to the curve \(y=\frac {xf(x)}{g(x)}\) at \(x = 4\) is given by \( \bigg [ \frac {d}{dx}\frac {xf(x)}{g(x)} \bigg ]_{x=4}\).
By the QuotientRule and the Product Rule, this derivative is given by \begin{align*} \bigg [ \frac {d}{dx}\frac {xf(x)}{g(x)} \bigg ]_{x=4} &=\bigg [ \frac {(xf'(x)+f(x))g(x)-xf(x)g'(x)}{\left (g(x)\right )^2} \bigg ]_{x=4}\\ &=\frac {(4f'(4)+f(4))g(4)-4f(4)g'(4)}{\left (g(4)\right )^2}\\ &=\frac {(4\cdot 5+3)(-2)-4\cdot 3\cdot 2}{\left (-2\right )^2}\\ &=\frac {(4\cdot 5+3)(-2)-4\cdot 3\cdot 2}{\left (-2\right )^2}\\ &=\frac {-35}{2}\\ \end{align*}