Trigonometric derivatives, the chain rule and implicit differentiation

Last week you differentiated polynomial and rational functions. This week the toolbox is completed: trigonometric functions, nested functions and curves where $y$ cannot be written explicitly. The chain rule is everywhere for the rest of the semester: logarithmic differentiation, related rates and substitution in integrals are all built on it. This week, make these four automatic:

1. Memorize the six trigonometric derivatives. The ones starting with “c” have a minus sign in the derivative.
2. The chain rule works from the outside in. Take the derivative of each layer, leave its inside as it is, and multiply them all.
3. $\sin^3(2x)$ has three layers: cube, sine, $2x$.
4. In implicit differentiation every term containing $y$ is multiplied by $y'$, because $y$ is actually a function of $x$.

Derivatives of the trigonometric functions

Everything comes from a single limit. If you put the addition formula $\sin(x+h)=\sin x\cos h+\cos x\sin h$ into the definition:

$\frac{\sin(x+h)-\sin x}{h}=\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}$

As $h\to0$, $\dfrac{\cos h-1}{h}\to0$ and $\dfrac{\sin h}{h}\to1$. The result is $(\sin x)'=\cos x$. Cosine follows the same way, and the other four come from the quotient rule.

Example 1 (derive the $\tan$ derivative yourself).

$(\tan x)'=\left(\frac{\sin x}{\cos x}\right)'=\frac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2 x}=\sec^2 x$

Example 2. $y=x^3\cos x$: $y'=3x^2\cos x-x^3\sin x$.

Example 3. $y=\dfrac{\sin x}{1+\cos x}$.

$y'=\frac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x)^2}=\frac{\cos x+\cos^2x+\sin^2x}{(1+\cos x)^2}=\frac{1+\cos x}{(1+\cos x)^2}=\frac{1}{1+\cos x}$

The identity $\sin^2x+\cos^2x=1$ is the key to simplifying trigonometric derivatives. Look for it before you leave the answer.

PITFALL · degrees and derivatives. The formulas are for radians. If $x$ were in degrees, $\sin x^\circ=\sin\frac{\pi x}{180}$ and its derivative would be $\frac{\pi}{180}\cos x^\circ$. In a calculus exam, angles are in radians.

The chain rule

If $y=f(g(x))$, then

$\frac{dy}{dx}=f'\big(g(x)\big)\cdot g'(x)$

In Leibniz notation, with $u=g(x)$:

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Intuition: if $u$ changes 3 times as fast as $x$, and $y$ changes 2 times as fast as $u$, then $y$ changes $2\cdot3=6$ times as fast as $x$.

Separating the layers

In the chain rule the real work is seeing the layers of the function. Ask yourself “what is the last operation performed?” That is the outer function.

Note the difference between $\sin(x^2)$ and $\sin^2 x$. In the first, you square first and then apply the sine; in the second, sine first, then square.

Example 4 (three layers). $y=\cos^3(4x^2-1)$.

layer 1:  u^3            derivative  3u^2         -> 3 cos^2(4x^2 - 1)
layer 2:  cos v          derivative  -sin v       -> -sin(4x^2 - 1)
layer 3:  4x^2 - 1       derivative  8x
multiply: y' = 3 cos^2(4x^2 - 1) · (-sin(4x^2 - 1)) · 8x
              = -24x cos^2(4x^2 - 1) sin(4x^2 - 1)

Example 5 (chain + product). $y=x^2\sqrt{1-x}$.

$y'=2x\sqrt{1-x}+x^2\cdot\frac{-1}{2\sqrt{1-x}}=\frac{4x(1-x)-x^2}{2\sqrt{1-x}}=\frac{4x-5x^2}{2\sqrt{1-x}}$

First the product rule, then the chain rule inside each piece where needed. To simplify, use the common denominator $2\sqrt{1-x}$.

PITFALL · forgetting to differentiate the inside. Writing $(\sin 5x)'=\cos 5x$ is the most common way to lose points on an exam. The correct answer is $5\cos 5x$. At every chain step ask “what is the derivative of the inside?”

The chain rule with table values

If a table of values is given instead of formulas, the same rule applies: for $h(x)=f(g(x))$, $h'(a)=f'(g(a))\cdot g'(a)$. $f'$ is evaluated at $g(a)$, not at $a$. First find $g(a)$, then look up that value in the $f'$ table.

Implicit differentiation

In the circle $x^2+y^2=25$ you cannot write $y$ in terms of $x$ with a single formula ($y=\pm\sqrt{25-x^2}$, two pieces). In implicit differentiation you differentiate without solving for $y$.

Method:

1. Differentiate both sides of the equation with respect to $x$.
2. In every term containing $y$ use the chain rule: $\dfrac{d}{dx}[y^n]=ny^{n-1}y'$.
3. Collect the terms containing $y'$ on one side, factor out $y'$, and divide.

Example 6 (circle). $x^2+y^2=25$, tangent at the point $(3,-4)$.

$2x+2yy'=0 \Rightarrow y'=-\frac xy$

At $(3,-4)$, $y'=-\dfrac{3}{-4}=\dfrac34$. Tangent: $y+4=\dfrac34(x-3)$. Geometry says the same: the tangent is perpendicular to the radius; the slope of the radius is $-\tfrac43$ and that of the tangent is $\tfrac34$.

Example 7 (with a product). $x^2y+y^3=10$, $y'$ at $(3,1)$.

d/dx[x^2 y] = 2x y + x^2 y'        (product rule)
d/dx[y^3]   = 3y^2 y'              (chain rule)
2xy + x^2 y' + 3y^2 y' = 0
y'(x^2 + 3y^2) = -2xy
y' = -2xy / (x^2 + 3y^2)
(3,1):  y' = -6 / 12 = -1/2

The second derivative by implicit differentiation

For $y''$, differentiate the $y'$ expression again. Replace the resulting $y'$ with the formula you found, then simplify using the curve equation.

Example 8. For $x^2+y^2=25$, $y'=-\dfrac xy$.

$y''=-\frac{1\cdot y-x\cdot y'}{y^2}=-\frac{y-x\left(-\frac xy\right)}{y^2}=-\frac{y^2+x^2}{y^3}=-\frac{25}{y^3}$

In the last step $x^2+y^2=25$ was used. At $(3,4)$, $y''=-\dfrac{25}{64}<0$: the upper semicircle is concave down, which makes sense.

The rational power rule

Implicit differentiation is also the proof of the rule $(x^{p/q})'=\frac pq x^{p/q-1}$: if $y=x^{p/q}$, then $y^q=x^p$; differentiate: $qy^{q-1}y'=px^{p-1}$, and solve for $y'$. For example $(x^{2/3})'=\frac23x^{-1/3}$.

PITFALL · differentiating $y$ as if it were a constant. Writing $\dfrac{d}{dx}[y^2]=2y$ is wrong; the correct answer is $2yy'$. Likewise $\dfrac{d}{dx}[xy]$ is not $y$, it is $y+xy'$. Every time $y$ appears, a $y'$ is born.

Common mistakes