Derivatives and the Rules of Differentiation, Worked Solutions

Ques­tion 1: Deriv­a­tives from first prin­ci­ples#

The prob­lem#

Find the deriv­a­tive from first prin­ci­ples of

(a) $x^2 + 3x$; (b) $\sqrt{x}$; (c) $\cos x$.

Under­stand­ing the prob­lem#

"From first prin­ci­ples" means you may not use the ready-made rules such as $\displaystyle \tfrac{d}{dx}x^n = nx^{n-1}$. You must work directly from the def­i­n­i­tion of the deriv­a­tive and eval­u­ate the limit.

The idea#

The def­i­n­i­tion is

$$\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}.$$

In each part we write $f(x + h) - f(x)$, sim­plify it until a fac­tor $h$ can­cels with the $h$ in the denom­i­na­tor, and only then let $h \to 0$. Putting $h = 0$ too early gives $\displaystyle \tfrac{0}{0}$, which is mean­ing­less.

Step-by-step solu­tion#

Part (a)

Step 1. Write $f(x + h)$ for $f(x) = x^2 + 3x$.

$$f(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h.$$

Step 2. Sub­tract $f(x)$; the $x^2$ and $3x$ terms can­cel.

$$f(x + h) - f(x) = 2xh + h^2 + 3h = h(2x + h + 3).$$

Step 3. Divide by $h$ (allowed, because $h \ne 0$ while we approach the limit) and let $h \to 0$.

$$\displaystyle f'(x) = \lim_{h \to 0}\frac{h(2x + h + 3)}{h} = \lim_{h \to 0}(2x + h + 3) = 2x + 3.$$

Part (b)

Step 1. Form the dif­fer­ence quo­tient for $f(x) = \sqrt{x}$ ($x > 0$).

$$\displaystyle \frac{\sqrt{x + h} - \sqrt{x}}{h}.$$

If we put $h = 0$ now we get $\displaystyle \tfrac{0}{0}$, so we must sim­plify first.

Step 2. Mul­ti­ply the top and bot­tom by the con­ju­gate $\sqrt{x + h} + \sqrt{x}$, and use $(p - q)(p + q) = p^2 - q^2$.

$$\displaystyle \frac{\sqrt{x + h} - \sqrt{x}}{h}\cdot\frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \frac{h}{h\left(\sqrt{x + h} + \sqrt{x}\right)}.$$

Step 3. Can­cel $h$ and take the limit.

$$\displaystyle f'(x) = \lim_{h \to 0}\frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}.$$

Part (c)

Step 1. Form the dif­fer­ence quo­tient for $f(x) = \cos x$.

$$\displaystyle \frac{\cos(x + h) - \cos x}{h}.$$

Step 2. Use the iden­tity $\displaystyle \cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2}$ with $C = x + h$, $D = x$.

$$\displaystyle \cos(x + h) - \cos x = -2\sin\left(x + \tfrac{h}{2}\right)\sin\tfrac{h}{2}.$$

Step 3. Rearrange so that the stan­dard limit $\displaystyle \lim_{\theta \to 0}\tfrac{\sin\theta}{\theta} = 1$ appears, with $\displaystyle \theta = \tfrac{h}{2}$.

$$\displaystyle \frac{\cos(x + h) - \cos x}{h} = -\sin\left(x + \tfrac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2}.$$

Step 4. Let $h \to 0$: the first fac­tor tends to $\sin x$ and the sec­ond to $1$.

$$\displaystyle \frac{d}{dx}\cos x = -\sin x \cdot 1 = -\sin x.$$

Check­ing the answer#

The rules you learn later give the same results: $\displaystyle \tfrac{d}{dx}(x^2 + 3x) = 2x + 3$ and $\displaystyle \tfrac{d}{dx}x^{1/2} = \tfrac12 x^{-1/2} = \tfrac{1}{2\sqrt{x}}$. For (c), at $x = 0$ the cosine graph is at its peak, so its slope should be $0$, and $-\sin 0 = 0$.

Answer#

(a) $2x + 3$; (b) $\displaystyle \frac{1}{2\sqrt{x}}$; (c) $-\sin x$.

Com­mon mis­take to avoid#

Sub­sti­tut­ing $h = 0$ before can­celling. The fac­tor $h$ must dis­ap­pear from the denom­i­na­tor first.

Ques­tion 2: Value of a deriv­a­tive at a point#

The prob­lem#

Find $f'(2)$ for $f(x) = 5x^3 - 4x$.

Under­stand­ing the prob­lem#

$f'(2)$ is the slope of the graph of $f$ at $x = 2$. First find the deriv­a­tive func­tion $f'(x)$, then sub­sti­tute $x = 2$.

The idea#

Use the power rule $\displaystyle \tfrac{d}{dx}x^n = nx^{n-1}$ together with the con­stant-mul­ti­ple and dif­fer­ence rules.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate term by term.

$$f'(x) = 5 \cdot 3x^2 - 4 \cdot 1 = 15x^2 - 4.$$

Step 2. Sub­sti­tute $x = 2$.

$$f'(2) = 15(4) - 4 = 60 - 4 = 56.$$

Check­ing the answer#

Esti­mate with a small step. $f(2) = 40 - 8 = 32$ and $f(2.001) \approx 32.05603$, so $\displaystyle \tfrac{f(2.001) - f(2)}{0.001} \approx 56.03$, very close to $56$.

Answer#

$f'(2) = 56$.

Com­mon mis­take to avoid#

Sub­sti­tut­ing $x = 2$ into $f$ first and then dif­fer­en­ti­at­ing the num­ber $32$ gives $0$. Always dif­fer­en­ti­ate first, then sub­sti­tute.

Ques­tion 3: Using the power, sum and prod­uct rules#

The prob­lem#

Dif­fer­en­ti­ate

(a) $x^{-3} + 2x^{1/2}$; (b) $(2x + 1)(x^2 - 4)$; (c) $(x - 1)^3$.

Under­stand­ing the prob­lem#

Three func­tions built from pow­ers of $x$. Part (a) has neg­a­tive and frac­tional pow­ers; (b) is a prod­uct; (c) is a cube of a bracket.

The idea#

The power rule $\displaystyle \tfrac{d}{dx}x^n = nx^{n-1}$ works for every real $n$, includ­ing neg­a­tive and frac­tional. For (b) use the prod­uct rule $(uv)' = u'v + uv'$. For (c), write the cube as a prod­uct (or expand it), since only the rules of this les­son are avail­able.

Step-by-step solu­tion#

Part (a)

Step 1. Dif­fer­en­ti­ate $x^{-3}$: $n = -3$, so the new power is $-4$.

$$\displaystyle \frac{d}{dx}x^{-3} = -3x^{-4}.$$

Step 2. Dif­fer­en­ti­ate $2x^{1/2}$: $\displaystyle n = \tfrac12$, so the new power is $\displaystyle -\tfrac12$.

$$\displaystyle \frac{d}{dx}2x^{1/2} = 2 \cdot \tfrac{1}{2}x^{-1/2} = x^{-1/2}.$$

Step 3. Add.

$$\displaystyle \frac{d}{dx}\left(x^{-3} + 2x^{1/2}\right) = -3x^{-4} + x^{-1/2} = -\frac{3}{x^4} + \frac{1}{\sqrt{x}}.$$

Part (b)

Step 1. Let $u = 2x + 1$ and $v = x^2 - 4$. Then $u' = 2$ and $v' = 2x$.

Step 2. Apply the prod­uct rule.

$$\displaystyle \frac{d}{dx}(uv) = u'v + uv' = 2(x^2 - 4) + (2x + 1)(2x).$$

Step 3. Expand and col­lect.

$$2x^2 - 8 + 4x^2 + 2x = 6x^2 + 2x - 8.$$

Part (c)

Step 1. Expand the cube using $(p - q)^3 = p^3 - 3p^2q + 3pq^2 - q^3$.

$$(x - 1)^3 = x^3 - 3x^2 + 3x - 1.$$

Step 2. Dif­fer­en­ti­ate term by term.

$$3x^2 - 6x + 3.$$

Step 3. Fac­tor to see the neat form.

$$3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2.$$

Check­ing the answer#

For (b), expand first: $(2x + 1)(x^2 - 4) = 2x^3 + x^2 - 8x - 4$, whose deriv­a­tive is $6x^2 + 2x - 8$. The same.

Answer#

(a) $-3x^{-4} + x^{-1/2}$; (b) $6x^2 + 2x - 8$; (c) $3(x - 1)^2$.

Ques­tion 4: Using the quo­tient and prod­uct rules#

The prob­lem#

Dif­fer­en­ti­ate

(a) $\displaystyle \frac{x^2}{x + 3}$; (b) $\displaystyle \frac{1 + \sin x}{\cos x}$; (c) $x\tan x$.

Under­stand­ing the prob­lem#

(a) and (b) are quo­tients of two func­tions; (c) is a prod­uct.

The idea#

Quo­tient rule: $\displaystyle \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, "deriv­a­tive of top times bot­tom, minus top times deriv­a­tive of bot­tom, over bot­tom squared". Prod­uct rule: $(uv)' = u'v + uv'$. We also need $\displaystyle \tfrac{d}{dx}\sin x = \cos x$, $\displaystyle \tfrac{d}{dx}\cos x = -\sin x$ and $\displaystyle \tfrac{d}{dx}\tan x = \sec^2 x$.

Step-by-step solu­tion#

Part (a)

Step 1. $u = x^2$, $u' = 2x$; $v = x + 3$, $v' = 1$.

Step 2. Apply the quo­tient rule.

$$\displaystyle \frac{d}{dx}\frac{x^2}{x + 3} = \frac{2x(x + 3) - x^2 \cdot 1}{(x + 3)^2}.$$

Step 3. Sim­plify the numer­a­tor: $2x^2 + 6x - x^2 = x^2 + 6x$.

$$\displaystyle = \frac{x^2 + 6x}{(x + 3)^2}.$$

Part (b)

Step 1. $u = 1 + \sin x$, $u' = \cos x$; $v = \cos x$, $v' = -\sin x$.

Step 2. Apply the quo­tient rule. Be care­ful with the minus sign in front of $uv'$.

$$\displaystyle \frac{\cos x \cdot \cos x - (1 + \sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin x + \sin^2 x}{\cos^2 x}.$$

Step 3. Use $\sin^2 x + \cos^2 x = 1$ in the numer­a­tor.

$$\displaystyle = \frac{1 + \sin x}{\cos^2 x}.$$

Part (c)

Step 1. $u = x$, $u' = 1$; $v = \tan x$, $v' = \sec^2 x$.

Step 2. Apply the prod­uct rule.

$$\displaystyle \frac{d}{dx}(x\tan x) = 1 \cdot \tan x + x\sec^2 x = \tan x + x\sec^2 x.$$

Check­ing the answer#

For (b), check at $x = 0$ numer­i­cally. The func­tion is $\displaystyle g(x) = \tfrac{1 + \sin x}{\cos x}$; $\displaystyle g(0.001) \approx \tfrac{1.001}{0.9999995} \approx 1.0010005$ and $g(0) = 1$, so the slope is about $1.0$. The for­mula gives $\displaystyle \tfrac{1 + 0}{1} = 1$. Cor­rect.

Answer#

(a) $\displaystyle \frac{x^2 + 6x}{(x + 3)^2}$; (b) $\displaystyle \frac{1 + \sin x}{\cos^2 x}$; (c) $\tan x + x\sec^2 x$.

Com­mon mis­take to avoid#

Writ­ing $uv' - u'v$ in the numer­a­tor (the wrong order) flips the sign of the answer. The quo­tient rule starts with $u'v$.

Ques­tion 5: Deriv­a­tives of trigono­met­ric com­bi­na­tions#

The prob­lem#

Dif­fer­en­ti­ate

(a) $\sin^2 x$ (treat­ing it as $\sin x \cdot \sin x$); (b) $\sec x - \tan x$.

Under­stand­ing the prob­lem#

In (a) you are told to treat the square as a prod­uct of two equal fac­tors, so that the prod­uct rule can be used. In (b) you need the stan­dard deriv­a­tives of $\sec x$ and $\tan x$.

The idea#

(a) Prod­uct rule with $u = v = \sin x$. (b) Dif­fer­ence rule with $\displaystyle \tfrac{d}{dx}\sec x = \sec x\tan x$ and $\displaystyle \tfrac{d}{dx}\tan x = \sec^2 x$.

Step-by-step solu­tion#

Part (a)

Step 1. $u = \sin x$, $v = \sin x$, so $u' = v' = \cos x$.

Step 2. Prod­uct rule.

$$\displaystyle \frac{d}{dx}(\sin x \cdot \sin x) = \cos x\sin x + \sin x\cos x = 2\sin x\cos x.$$

(This is also $\sin 2x$.)

Part (b)

Step 1. Dif­fer­en­ti­ate each term.

$$\displaystyle \frac{d}{dx}\sec x = \sec x\tan x, \qquad \frac{d}{dx}\tan x = \sec^2 x.$$

Step 2. Sub­tract.

$$\displaystyle \frac{d}{dx}(\sec x - \tan x) = \sec x\tan x - \sec^2 x.$$

This can also be writ­ten $\sec x(\tan x - \sec x)$.

Check­ing the answer#

For (a), $\displaystyle \sin^2 x = \tfrac{1 - \cos 2x}{2}$. Its deriv­a­tive, using $\displaystyle \tfrac{d}{dx}\cos 2x = -2\sin 2x$, is $\sin 2x = 2\sin x\cos x$. Agrees.

Answer#

(a) $2\sin x\cos x$; (b) $\sec x\tan x - \sec^2 x$.

Ques­tion 6: Slope of a tan­gent and hor­i­zon­tal tan­gents#

The prob­lem#

Find the slope of the tan­gent to $y = x^3 - 3x + 1$ at $x = 2$. At which points is the tan­gent hor­i­zon­tal?

Under­stand­ing the prob­lem#

The slope of the tan­gent at a point is the value of $\displaystyle \tfrac{dy}{dx}$ there. A hor­i­zon­tal tan­gent has slope $0$.

The idea#

Dif­fer­en­ti­ate, sub­sti­tute $x = 2$ for the first part, and solve $\displaystyle \tfrac{dy}{dx} = 0$ for the sec­ond.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$\displaystyle \frac{dy}{dx} = 3x^2 - 3.$$

Step 2. Slope at $x = 2$.

$$3(2)^2 - 3 = 12 - 3 = 9.$$

Step 3. Hor­i­zon­tal tan­gent: set the slope to zero.

$$3x^2 - 3 = 0 \;\Longrightarrow\; x^2 = 1 \;\Longrightarrow\; x = 1 \text{ or } x = -1.$$

Step 4. Find the points on the curve. At $x = 1$: $y = 1 - 3 + 1 = -1$. At $x = -1$: $y = -1 + 3 + 1 = 3$.

Check­ing the answer#

$y$ at $x = 0.9$, $1$, $1.1$ is $-0.971$, $-1$, $-0.969$: the curve bot­toms out near $x = 1$, con­sis­tent with a hor­i­zon­tal tan­gent there.

Answer#

Slope $9$ at $x = 2$; the tan­gent is hor­i­zon­tal at $x = \pm 1$, i.e. at the points $(1, -1)$ and $(-1, 3)$.

Ques­tion 7: Rate of change of the area of a cir­cle#

The prob­lem#

The area of a cir­cle is $A = \pi r^2$. How fast does $A$ change with respect to $r$ when $r = 5$?

Under­stand­ing the prob­lem#

"How fast $A$ changes with $r$" is the deriv­a­tive $\displaystyle \tfrac{dA}{dr}$, the instan­ta­neous rate of change of area per unit increase in radius. We need its value at $r = 5$.

The idea#

Dif­fer­en­ti­ate $\pi r^2$ with respect to $r$; $\pi$ is a con­stant mul­ti­plier.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$\displaystyle \frac{dA}{dr} = \pi \cdot 2r = 2\pi r.$$

Step 2. Sub­sti­tute $r = 5$.

$$\displaystyle \frac{dA}{dr}\Big|_{r = 5} = 10\pi \approx 31.4.$$

Check­ing the answer#

$2\pi r$ is the cir­cum­fer­ence: increas­ing the radius a lit­tle adds a thin ring whose area is about cir­cum­fer­ence $\times$ thick­ness. That makes sense.

Answer#

$\displaystyle \frac{dA}{dr} = 10\pi$ square units per unit of radius when $r = 5$.

Ques­tion 8: Deriv­a­tive of a long poly­no­mial at x = 1#

The prob­lem#

If $f(x) = 1 + x + x^2 + \cdots + x^{50}$, find $f'(1)$.

Under­stand­ing the prob­lem#

$f$ is a sum of $51$ terms, the pow­ers of $x$ from $0$ to $50$. We want the deriv­a­tive at $x = 1$.

The idea#

Dif­fer­en­ti­ate term by term with the power rule, then put $x = 1$; every power of $1$ is $1$, leav­ing a sum of whole num­bers.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate each term: $\displaystyle \tfrac{d}{dx}1 = 0$, $\displaystyle \tfrac{d}{dx}x^k = kx^{k-1}$.

$$f'(x) = 0 + 1 + 2x + 3x^2 + \cdots + 50x^{49}.$$

Step 2. Put $x = 1$.

$$f'(1) = 1 + 2 + 3 + \cdots + 50.$$

Step 3. Use the sum of the first $n$ nat­ural num­bers, $\displaystyle \tfrac{n(n + 1)}{2}$, with $n = 50$.

$$\displaystyle f'(1) = \frac{50 \times 51}{2} = 1275.$$

Check­ing the answer#

For a small case, $f(x) = 1 + x + x^2$ gives $\displaystyle f'(1) = 1 + 2 = 3 = \tfrac{2 \cdot 3}{2}$. The pat­tern holds.

Answer#

$f'(1) = 1275$.

Ques­tion 9: Deriv­a­tive of tan x from the quo­tient rule#

The prob­lem#

Derive $\displaystyle \frac{d}{dx}\tan x = \sec^2 x$ using the quo­tient rule.

Under­stand­ing the prob­lem#

This is a short proof. You must show, using only $\displaystyle \tfrac{d}{dx}\sin x = \cos x$, $\displaystyle \tfrac{d}{dx}\cos x = -\sin x$ and the quo­tient rule, that the deriv­a­tive of $\tan x$ is $\sec^2 x$.

The idea#

Write $\displaystyle \tan x = \frac{\sin x}{\cos x}$, apply the quo­tient rule, and sim­plify with $\sin^2 x + \cos^2 x = 1$.

Step-by-step solu­tion#

Step 1. Express $\tan x$ as a quo­tient: $u = \sin x$, $v = \cos x$, with $u' = \cos x$, $v' = -\sin x$.

Step 2. Apply the quo­tient rule.

$$\displaystyle \frac{d}{dx}\tan x = \frac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}.$$

Step 3. The numer­a­tor is $1$ by the Pythagorean iden­tity.

$$\displaystyle = \frac{1}{\cos^2 x} = \sec^2 x.$$

This holds wher­ever $\cos x \ne 0$. $\blacksquare$

Check­ing the answer#

At $x = 0$, $\tan x$ behaves like $x$ (slope $1$), and $\sec^2 0 = 1$. At $\displaystyle x = \tfrac{\pi}{4}$, $\sec^2 = 2$, and indeed $\tan$ is steeper there.

Answer#

$\displaystyle \frac{d}{dx}\tan x = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$.

Ques­tion 10: Deriv­a­tive of a gen­eral lin­ear frac­tion#

The prob­lem#

Find $\displaystyle \frac{d}{dx}\left(\frac{ax + b}{cx + d}\right)$, where $a, b, c, d$ are con­stants.

Under­stand­ing the prob­lem#

This is a quo­tient of two lin­ear expres­sions. The let­ters $a, b, c, d$ are fixed num­bers, not vari­ables. The result is a use­ful gen­eral for­mula (Exam­ple 6 is the spe­cial case $a = b = c = 1$, $d = -1$).

The idea#

Quo­tient rule with $u = ax + b$ and $v = cx + d$.

Step-by-step solu­tion#

Step 1. $u' = a$, $v' = c$.

Step 2. Apply the quo­tient rule.

$$\displaystyle \frac{a(cx + d) - (ax + b)c}{(cx + d)^2}.$$

Step 3. Expand the numer­a­tor: $acx + ad - acx - bc$. The $acx$ terms can­cel.

$$\displaystyle = \frac{ad - bc}{(cx + d)^2}.$$

Check­ing the answer#

With $a = b = c = 1$, $d = -1$ this gives $\displaystyle \tfrac{-1 - 1}{(x - 1)^2} = -\tfrac{2}{(x - 1)^2}$, exactly Exam­ple 6 of the les­son. Also, if $ad - bc = 0$ the frac­tion is a con­stant, and its deriv­a­tive is $0$, as the for­mula says.

Answer#

$\displaystyle \frac{d}{dx}\left(\frac{ax + b}{cx + d}\right) = \frac{ad - bc}{(cx + d)^2}$.