Modules › Differentiation rules, implicit differentiation, related rates, linearization
Differentiation rules, implicit differentiation, related rates, linearization
Rules of differentiation, chain rule, implicit differentiation, related rates and linear approximation.
Lecture 4 · Tutorial 4 · Derivatives
Textbook sections: Thomas ch. 3 Derivatives (3.1–3.11) · Stewart ch. 2 (2.7–2.8) and ch. 3 Differentiation Rules (3.1–3.5, 3.9, 3.10)
Textbook sections: Thomas ch. 3 Derivatives (3.1–3.11) · Stewart ch. 2 (2.7–2.8) and ch. 3 Differentiation Rules (3.1–3.5, 3.9, 3.10)
Differentiation rules, implicit differentiation, related rates, linearization
Calculus I · Module 5
After this module you can
- Differentiate products, quotients and composites quickly and without mistakes.
- Differentiate implicitly, also twice.
- Solve related-rates problems with a standard procedure.
- Use the linearization $L(x)=f(a)+f'(a)(x-a)$ and differentials to estimate.
Rules
| Rule | Formula |
|---|---|
| power | $(x^n)'=nx^{n-1}$ |
| product | $(uv)'=u'v+uv'$ |
| quotient | $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$ |
| chain | $(f(g(x)))'=f'(g(x))\,g'(x)$ |
| trig | $(\sin x)'=\cos x$, $(\cos x)'=-\sin x$, $(\tan x)'=\sec^2x$, $(\sec x)'=\sec x\tan x$ |
Chain rule: outer–inner
- $y=\sin\left(\frac{t}{\sqrt{t+1}}\right)$: derivative of the outer $\sin$ is $\cos(\cdot)$, times the derivative of the inner $\frac{t}{\sqrt{t+1}}$.
- $y=(1+\tan^4\frac t{12})^3$: three layers: cube, fourth power of $\tan$, argument $\frac{t}{12}$.
- Simplify before differentiating when possible: $\frac{\sin\theta+\cos\theta}{\cos\theta}=\tan\theta+1$.
- Check by differentiating a simpler equivalent form.
Implicit differentiation
- Differentiate both sides with respect to $x$, treating $y=y(x)$: $\frac{d}{dx}y^n=ny^{n-1}y'$.
- Solve for $y'$. For $F(x,y)=0$: $y'=-\frac{F_x}{F_y}$.
- Second derivative: differentiate $y'$ again and substitute $y'$.
- Example: $x^3+y^3=16$ at $(2,2)$: $y'=-\frac{x^2}{y^2}=-1$ and $y''=-2$.
Related rates: the procedure
- 1. Draw a picture, name the variables that change with time $t$.
- 2. Write one equation relating them (Pythagoras, area, $V=IR$, …).
- 3. Differentiate with respect to $t$ (chain rule!).
- 4. Substitute the instant's values and solve. Example: $l'=-2$, $w'=2$ at $l=12,w=5$: $A'=l'w+lw'=14$, diagonal $d'=-\frac{14}{13}$.
Linearization and differentials
- $L(x)=f(a)+f'(a)(x-a)\approx f(x)$ near $a$; differential $dy=f'(x)\,dx$.
- $(1+x)^k\approx1+kx$: $(1.0002)^{50}\approx1.01$, $\sqrt[3]{1.009}\approx1.003$.
- $\sin x$ at $a=0$: $L(x)=x$; at $a=\pi$: $L(x)=\pi-x$.
- Area of a circle: $dA=2\pi r\,dr$; $r:2.00\to2.02$ gives $dA\approx0.08\pi$ (a $2\%$ change).
Typical mistakes
Missing chain factor$\frac{d}{dx}\sin(x^2)=2x\cos(x^2)$, not $\cos(x^2)$.
Product rule$(uv)'\neq u'v'$.
Related ratesDifferentiate first, substitute numbers after.
Practice next
- Chapter quizzes Ch. 2 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Plot and derivative
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Differentiation rules (Computation problems 3, Q7(1))
Find $f'$ for $f=\dfrac{\sqrt s-1}{\sqrt s+1}$.
Show the answer and the solution
Answer: $\frac{1}{\sqrt{s} \left(\sqrt{s} + 1\right)^{2}}$
Quotient rule: $\frac{(\sqrt s+1)\frac1{2\sqrt s}-(\sqrt s-1)\frac1{2\sqrt s}}{(\sqrt s+1)^2}$. $f'=\frac{1}{\sqrt{s} \left(\sqrt{s} + 1\right)^{2}}$.
Worked example 2 · Higher derivatives (Computation problems 4, Q5)
Find $y''$ for $y=\left(1+\dfrac1x\right)^3$.
Show the answer and the solution
Answer: $\frac{6 \left(x + 1\right) \left(x + 2\right)}{x^{5}}$
$y'=3(1+x^{-1})^2(-x^{-2})$, then differentiate again with the product rule: $y''=\frac{6 \left(x + 1\right) \left(x + 2\right)}{x^{5}}$.
Worked example 3 · Implicit differentiation (Computation problems 4, Q6(a))
Use implicit differentiation to find $\dfrac{dy}{dx}$ for $x^2y+xy^2=6$.
Show the answer and the solution
Answer: $- \frac{y \left(2 x + y\right)}{x \left(x + 2 y\right)}$
Differentiate both sides w.r.t. $x$ treating $y=y(x)$, then solve for $y'=-\frac{F_x}{F_y}$: $\frac{dy}{dx}=- \frac{y \left(2 x + y\right)}{x \left(x + 2 y\right)}$.
Worked example 4 · Related rates (Tutorial 4, Q7(d))
In $V=IR$, $V$ increases at $1$ V/s and $I$ decreases at $\frac13$ A/s. When $V=12$ V and $I=2$ A, $\dfrac{dR}{dt}=$ (ohms/s)
Show the answer and the solution
Answer: $\frac{3}{2}$
$R=V/I$; $R'=\frac{V'I-VI'}{I^2}=\frac{1\cdot2-12\cdot(-\frac13)}{4}=\frac{6}{4}=\frac32>0$: $R$ is increasing.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Plot and derivative