18.01: Single Variable Calculus
Section 3
Derivatives of Products, Quotients, Sine, and Cosine
Massachusetts Institute of Technology
Last Edit Date: 06/05/2024
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Derivative Formulas¶
There are two kinds of derivative formulas:
Specific Examples: $\frac{d}{dx} x^n$ or $\frac{d}{dx} \left( \frac{1}{x} \right)$
General Examples: $(u + v)' = u' + v'$ and $(cu)' = cu'$ (where $c$ is a constant)
A notational convention we will use today is:
$$(u + v)(x) = u(x) + v(x);~~~uv(x) = u(x)v(x)$$
Proof of $(u + v) = u' + v'$. (General)¶
Start by using the definition of the derivative.
$$ \begin{aligned} (u + v)'(x) &= \lim_{\Delta x \rightarrow 0} \frac{(u + v)(x + \Delta x) - (u + v)(x)}{\Delta x} \\ &= \lim_{\Delta x \rightarrow 0} \frac{u(x + \Delta x) + v(x + \Delta x) - u(x) - v(x)}{\Delta x} \\ &= \lim_{\Delta x \rightarrow 0} \left\{ \frac{u(x + \Delta x) - u(x)}{\Delta x} + \frac{v(x + \Delta x) - v(x)}{\Delta x} \right\} \\ (u + v)'(x) &= u'(x) + v'(x) \end{aligned} $$
Follow the same procedure to prove that $(cu)' = cu'$.
Derivatives of $\sin x$ and $\cos x$. (Specific)¶
Last time, we computed
$$ \begin{aligned} \lim_{x \rightarrow 0} \frac{\sin x}{x} &= 1\\ \frac{d}{dx}(\sin x) |_{x = 0} &= \lim_{\Delta \rightarrow 0} \frac{\sin (0 + \Delta x) - sin(0)}{\Delta x} = \lim_{\Delta \rightarrow 0} \frac{\sin(\Delta x)}{\Delta x} = 1 \\ \frac{d}{dx}(\cos x) |_{x = 0} &= \lim_{\Delta \rightarrow 0} \frac{\cos (0 + \Delta x) - cos(0)}{\Delta x} = \lim_{\Delta \rightarrow 0} \frac{\cos(\Delta x) - 1}{\Delta x} = 0 \\ \end{aligned} $$
So, we know the value of $\frac{d}{dx} \sin x$ and of $\frac{d}{dx} \cos x$ at $x = 0$. Let us find these for arbitrary $x$.
$$\frac{d}{dx} \sin x = \lim_{\Delta x \rightarrow x} \frac{\sin (x + \Delta x) - \sin (x)}{\Delta x}$$
Recall:
$$\sin (a + b) = \sin(a) \cos(b) + \sin(b) \cos(a)$$
So,
$$ \begin{aligned} \frac{d}{dx} \sin x &= \lim_{\Delta x \rightarrow 0} \frac{\sin x \cos \Delta x + \cos x \sin \Delta x - \sin(x)}{\Delta x} \\ &= \lim_{\Delta \rightarrow 0} \left[ \frac{\sin x (\cos \Delta x - 1)}{\Delta x} + \frac{\cos x \sin \Delta x}{\Delta x} \right] \\ &= \lim_{\Delta \rightarrow 0} \sin x \left( \frac{\cos \Delta x - 1}{\Delta x} \right) + \lim_{\Delta \rightarrow 0} \cos x \left( \frac{\sin \Delta x }{\Delta x} \right) \\ \end{aligned} $$
Since $\frac{\cos \Delta x - 1}{\Delta x} \rightarrow 0$ and that $\frac{\sin \Delta x}{\Delta x} \rightarrow 1$, the equation above simplifies to
$$\frac{d}{dx} \sin x = \cos x$$
A similar calculation gives
$$\frac{d}{dx} \cos x = - \sin x$$
Product formula (General)¶
$$(uv)' = u'v + uv'$$
Proof:
$$(uv)' = \lim_{\Delta x \rightarrow 0} \frac{(uv)(x + \Delta x) - (uv)(x)}{\Delta x} = \lim_{\Delta \rightarrow 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x}$$
Now obviously,
$$u(x + \Delta x)v(x) - u(x + \Delta x)v(x) = 0$$
so adding that to the numerator will not change anything.
$$(uv)' = \lim_{\Delta \rightarrow 0} \frac{u(x + \Delta x)v(x) - u(x)v(x) + u(x + \Delta x)v(x + \Delta x) - u(x + \Delta x)v(x)}{\Delta x}$$
We can re-arrange that expression to get
$$(uv)' = \lim_{\Delta x \rightarrow 0} \left( \frac{u(x + \Delta x) - u(x)}{\Delta x} \right) v(x) + u(x + \Delta x) \left( \frac{v(x + \Delta x) - v(x)}{\Delta x} \right)$$
Note: we also used the fact that
$$\lim_{\Delta x \rightarrow 0} u(x + \Delta x) = u(x)~~~~~\text{true because}~u~\text{is continuous}$$
This proof of the product rule assumes that $u$ and $v$ have derivatives, which implies both functions are continuous.
An intuitive justification:
We want to find the difference in area between the large rectangle and the sammer, inner rectangle. The inner (orange) rectangle has area $w$. Define $\Delta u$, the change in $u$, by
$$\Delta u = u(x + \Delta x) - u(x)$$
We also abbreviate $u = u(x)$, so that $u(x + \Delta x) = u + \Delta u$, and, similarly, $v(x + \Delta x) = v + \Delta v$. Therefore the are of the largest rectangle is $(u + \Delta u)(v + \Delta v)$.
If you let $v$ increase and kee $u$ constant, you add the area shaded in red. If you let $u$ increase and keep $v$ constant, you add the area shaded in yellow. The sum of areas of the red and yello rectangles is:
$$[(u + \Delta u)(v + \Delta v) - uv] \approx u \Delta v + v\Delta u$$
(Divide by $\Delta x$ and let $\Delta x \rightarrow 0$ to finish the argument.)
Quotient formula (General)¶
To calculate the derivatice of $u / v$, we use the notations $\Delta u$ and $\Delta v$ above. Thus,
$$ \begin{aligned} \frac{u(x + \Delta x)}{v(x + \Delta x)} - \frac{u(x)}{v(x)} &= \frac{u + \Delta u}{v + \Delta v} - \frac{u}{v} \\ &= \frac{(u + \Delta u)v - u(v + \Delta v)}{(v + \Delta v) v}~~~~\text{(common denominator)}\\ &= \frac{(\Delta u)v - u(\Delta v)}{(v + \Delta v)v}~~~~\text{(cancel $uv - uv$)} \end{aligned} $$
Hence,
$$\frac{1}{\Delta x} \left( \frac{u + \Delta u}{v + \Delta v} - \frac{u}{v} \right) = \frac{\left( \frac{\Delta u}{\Delta x} \right)v - u\left( \frac{\Delta v}{\Delta x} \right)}{(v + \Delta v)v} \rightarrow \frac{v \left( \frac{du}{dx} \right) - u \left( \frac{dv}{dx} \right)}{v^2}~~~~~\text{as $\Delta x \rightarrow 0$}$$
Therefore,
$$\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$$
.