1 Derivative Rules

1.1 Solution 1

$f(x) = 2x^6 - 3x^4 + x^3 - 7x^2 + 4x-8$

$f'(x) = 12 - 12x^3 + 3x^2 - 14x + 4$

1.2 Solution 2

$g(x) = \dfrac{x^4}{2} - \dfrac{2x^3}{5} + x - 5$

$g'(x) = \dfrac{4x^3}{2} - \dfrac{6x^2}{5} + 1$

$= x^3 - \dfrac{6x^2}{5} + 1$

1.3 Solution 3

$f(x) = \dfrac{x^2+3x-5}{3x^2-5x+1}$

Using the quotient rule,

$f'(x) = \dfrac{(3x^2-5x+1)(2x+3)-(x^2+3x-5)(6x-5)}{(3x^2-5x+1)^2}$

$= \dfrac{6x^3+9x^2-10x^2-15x+2x+3-(6x^3-5x^2+18x^2-15x-30x+25)}{(3x^2-5x+1)^2}$

$= \dfrac{32x-14x^2-22}{(3x^2-5x+1)^2}$

1.4 Solution 4

Using the product rule,

$h'(x) = (x^3-4x)(3x^2-4) + (x^3-4x)(3x^2-4)$

$= 2(x^3-4x)(3x^2-4)$

1.5 Solution 5

$g(x) = \dfrac{2x+3}{x^3+2}$

$g'(x) = \dfrac{(x^3+2)(2)-(2x+3)(3x^2)}{(x^3+2)^2}$

$= \dfrac{2x^3+4-(6x^3+9x^2)}{(x^3+2)^2}$

$= \dfrac{4-9x^2-4x^3}{(x^3+2)^2}$

1.6 Solution 6

$f(x) = x^5 + 3\sqrt[5]{x}$

$f(x) = x^5 + 3(x)^{1/5}$

$f'(x) = 5x^4 + 3(1/5)x^{1/5 - 1}$

$= 5x^4 + \dfrac{3}{5}x^{-4/5}$

1.7 Solution 7

$g(x) = \dfrac{4x-1}{x^2 + \sqrt{x}}$

Using the quotient rule,

$g'(x) = \dfrac{(x^2+\sqrt{x})(4)-(4x-1)(2x+\dfrac{1}{2\sqrt{x}})}{(x^2 + \sqrt{x})^2}$

$= \dfrac{4x^2+4\sqrt{x}-(8x^2+2\sqrt{x}-2x-\dfrac{1}{2\sqrt{x}})}{(x^2 + \sqrt{x})^2}$

$= \dfrac{-4x^2+2\sqrt{x}+2x + \dfrac{1}{2\sqrt{x}}}{(x^2 + \sqrt{x})^2}$

1.8 Solution 8

$h(x) = \dfrac{3 + \dfrac{2}{x-1}}{x+1}$

Using the quotient rule,

$h'(x) = \dfrac{(x+1)\dfrac{d}{dx}(\dfrac{2}{x-1}) - (3 + \dfrac{2}{x-1})}{(x+1)^2}$

$\dfrac{d}{dx} (\dfrac{2}{x-1}) = \dfrac{(x-1).0 - 2.1}{(x-1)^2}$

$= \dfrac{-2}{(x-1)^2}$

$h'(x) = \dfrac{\dfrac{(x+1)(-2)}{(x-1)^2} - \dfrac{3(x-1)+2}{x-1}}{(x+1)^2}$

$= \dfrac{\dfrac{(x+1)(-2)}{(x-1)^2} - \dfrac{3x-1}{x-1}}{(x+1)^2}$

$= \dfrac{\dfrac{(x+1)(-2)-(3x-1)(x-1)}{(x-1)^2}}{(x+1)^2}$

$= \dfrac{(x+1)(-2)-(3x-1)(x-1)}{(x-1)^2(x+1)^2}$

1.9 Solution 9

$f(x) = \dfrac{\sqrt{x}-1}{\sqrt[3]{x}+1}$

Using the quotient rule,

$f'(x) = \dfrac{(\sqrt[3]{x}+1)(\dfrac{1}{x\sqrt{x}}) - (\sqrt{x}-1)(\dfrac{1}{3} - x^{-2/3})}{(\sqrt[3]{x} + 1)^2}$

$= \dfrac{\dfrac{\sqrt[3]{x}+1}{2\sqrt{x}} - \dfrac{\sqrt{x}-1}{3x^{2/3}}}{(\sqrt[3]{x} + 1)^2}$

$= \dfrac{3x^{2/3}x^{1/3}+3x^{2/3}-2x+2x^{1/2}}{6x^{1/2}x^{2/3}(\sqrt[3]{x} + 1)^2}$

$= \dfrac{3x^{1/6}x^{1/3}+3x^{1/6}-2x^{1/2}+2}{6x^{2/3}(\sqrt[3]{x}+1)^2}$

$= \dfrac{3\sqrt{x} + 3x^{1/6} - 2\sqrt{x} + 2}{6\sqrt[3]{x^2}(\sqrt[3]{x} + 1)^2}$

$= \dfrac{\sqrt{x} + 3\sqrt[6]{x} + 2}{6\sqrt[3]{x^2}(\sqrt[3]{x} + 1)^2}$

1.10 Solution 10

$h(x) = x^{7/3}$

$x^{7/3} = x^{2 + 1/3}$

$= x^2 . x^{1/3}$

$= x^2 . \sqrt[3]{x}$

$h(x) = x^2 \sqrt[3]{x}$

Using the product rule,

$h'(x) = x^2\dfrac{1}{3}x^{1/3 - 1} + \sqrt[3]{x}2x$

$= \dfrac{1}{3}x^2x^{-2/3} + 2x^{1/2}x^{1/3}$

$= \dfrac{1}{3}x^{4/3} + 2x^{5/6}$

1.11 Solution 11

$h(x) = x^{2/3}$

$x^{2/3} = (x^2)^{1/3} = (x^{1/3})^2 = \sqrt[3]{x}\sqrt[3]{x}$

Using the product rule,

$h'(x) = \sqrt[3]{x}\dfrac{1}{3}x^{-2/3} + \sqrt[3]{x}\dfrac{1}{3}x^{-2/3}$

$= \dfrac{2}{3}x^{1/3}x^{-2/3}$

$= \dfrac{2}{3}x^{-1/3}$

$= \dfrac{2}{3x^{1/3}}$

$= \dfrac{2}{3\sqrt[3]{x}}$

1.12 Solution 12

$f(x) = \sin x \cos x$

Using the product rule,

$f'(x) = \sin x (- \sin x) + \cos x \cos x$

$= -\sin^2 x + \cos^2 x$

$= \cos^2 x - \sin^2 x$

1.13 Solution 13

$f(x) = \sqrt{x} \sin x \cos x$

We will use product rule and also the result from solution 12,

$f'(x) = \sqrt{x}(\cos^2 x - \sin^2 x) + (\sin x \cos x \dfrac{1}{2}x^{-1/2})$

$= \sqrt{x}(\cos^2 x - \sin^2 x) + \dfrac{\sin x \cos x}{2\sqrt{x}}$

$= \dfrac{2x(\cos^2 x - \sin^2 x) + \sin x \cos x}{2\sqrt{x}}$

1.14 Solution 14

$g(x) = x^2 \tan x$

Using the product rule,

$g'(x) = x^2 \sec^2 x + \tan x (2x)$

$= x^2 \sec^2 x + 2x\tan x$

1.15 Solution 15

$h(x) = \dfrac{x}{\sec x + \tan x}$

Using the quotient rule,

$h'(x) = \dfrac{(\sec x + \tan x)-x(\sec x \tan x + \sec^2 x)}{(\sec x + \tan x)^2}$

$(\sec x + \tan x)-x(\sec x \tan x + \sec^2 x)$

$= \sec x + \tan x - x \sec x \tan x - x \sec^2 x$

$= \dfrac{1}{\cos x} + \dfrac{\sin x}{\cos x} - \dfrac{x \sin x}{\cos^2 x} - \dfrac{x}{\cos^2 x}$

$= \dfrac{\cos x}{\cos^2 x} + \dfrac{\sin x \cos x}{\cos^2 x} - \dfrac{x\sin x}{\cos^2 x} - \dfrac{x}{\cos^2 x}$

$= \dfrac{(\cos x + \sin x \cos x - x \sin x - x)}{\cos^2 x}$

$(\sec x + \tan x)^2 = (\dfrac{1}{\cos x} + \dfrac{\sin x}{\cos x})^2$

$= (\dfrac{1+\sin x}{\cos x})^2$

$= \dfrac{(1+\sin x)^2}{\cos^2 x}$

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

$= \dfrac{\cos x (1 + \sin x) - x (\sin x + 1)}{(1 + \sin x)^2}$

$= \dfrac{\cos x - x}{1 + \sin x}$

1.16 Solution 16

$f(x) = \dfrac{\sin x}{\cos x + \sec x}$

Using the quotient rule,

$f'(x) - \dfrac{(\cos x + \sec x)(\cos x)-(\sin x)(-\sin x + \sec x \tan x)}{(\cos x + \sec x)^2}$

$= \dfrac{(\cos^2 x + 1)-(-\sin^2 x + \tan^2 x)}{(\cos x + \sec x)^2}$

$= \dfrac{\cos^2 + 1 + \sin^2 x - \tan^2 x}{(\cos x + \sec x)^2}$

$= \dfrac{2-\tan^2 x}{(\cos x + \sec x)^2}$

1.17 Solution 17

$y = 3x - \dfrac{2}{x}$

$f(x) = 3x - \dfrac{2}{x}$

We know that $f'(a)$ is the rate of change of $f(x)$ with respect to $x$ at $x=a$ which is also the slope of the line tangent to the graph of $f$ at the point $(a, f(a))$.

$f'(x) = 3 - (-1).2.x^{-2}$

$= 3 + \dfrac{2}{x^2}$

Slope at $(2,5) = 3 + \dfrac{2}{2^2} = 3 + \dfrac{1}{2} = \dfrac{7}{2}$

Equation of line: $y = mx + c$

$y = \dfrac{7}{2}{x} + c$

$5 = \dfrac{7.2}{2} + c$

$c = -2$

$y = \dfrac{7x}{2} - 2$

1.18 Solution 18

Equation of the curve: $y = x^2$

$f(x) = x^2$

$f'(x) = 2x$

We know that $f'(a)$ is the rate of change of $f(x)$ with respect to $x$ at $x=a$ which is also the slope of the line tangent to the graph of $f$ at the point $(a,f(a))$

Let point $P = (x_1, y_1)$ and $Q = (x_2, y_2)$

So, $P = (x_1, x_1^2)$

$Q = (x_2, x_2^2)$

We know that the line tangent to the curve at $P$ and $Q$ passes through $(3,5)$.

Slope at $P = \dfrac{5-x_1^2}{3-x_1}$

We also know that $f'(x) = 2x$

$f'(x_1) = 2x_1$

$2x_1 = \dfrac{5-x_1^2}{3-x_1}$

$6x_1 - 2x_1^2 = 5 - x_1^2$

$6x_1 = 5 + x_1^2$

$5 + x_1^2 = 6x_1$

$x_1^2 - 6x_1 + 5 = 0$

$x_1^2 - 5x_1 - x_1 + 5 = 0$

$x_1(x_1 - 5) - 1(x_1 - 5) = 0$

$(x_1 - 5)(x_1 - 1) = 0$

$x_1 = 5$ or $x_1 = 1$

So, $P = (5, 25)$

$Q = (1,1)$

1.19 Solution 19

$y = \dfrac{1}{x}$

The tangent to the above curve at point $P$ passes through $(4,0)$.

$f(x) = \dfrac{1}{x}$

$f'(x) = -1x^{-1-1} = \dfrac{-1}{x^2}$

Let point $P = (x_1, y_1)$

$P = (x_1, \dfrac{1}{x_1})$

$m = \dfrac{0 - \dfrac{1}{x_1}}{4 - x_1}$

$= \dfrac{0-1}{x_1(4-x_1)} = \dfrac{-1}{x_1(4-x_1)}$

$\dfrac{-1}{x_1^2} = \dfrac{-1}{x_1(4-x_1)}$

$\dfrac{1}{x_1} = \dfrac{1}{4-x_1}$

$x_1 = 4 - x_1$

$2x_1 = 4$

$x_1 = 2$

$y_1 = \dfrac{1}{2}$

So the point $P$ is $(2,\dfrac{1}{2})$

1.20 Solution 20

Observer $= (0,0)$

Initial rocket position $= (10, 0)$

Equation of the hill $y = -x^2 + 10x - 16$

Domain of the above equation $[2,8]$

Let's assume that the rocket is at point $P$ when the observer first sees it.

Let $P = (10, y_1)$

The tangent of the equation of hill passes through $(0,0)$ and $(10,y_1)$.

Let $Q$ be the point in the curve such that it's tangent passes through $(0,0)$ and $(10, y_1)$.

$Q = (x_2, y_2)$

$f(x) = -x^2 + 10x -16$

$f'(x) = -2x + 10$

At $Q$, $f'(x_2) = -2x_2 + 10$

Also from the equation of the hill,

$-y_2 = -x_2^2 + 10x_2 - 16$

Also, $m = \dfrac{y_2 - 0}{x_2 - 0} = \dfrac{y_2}{x_2}$

So, $\dfrac{y_2}{x_2} = -2x_2 + 10$

$y_2 = -2x_2^2 + 10x_2$

From the previous equation,

$-2x_2^2 + 10x_2 = -x_2^2 + 10x_2 - 16$

$16 = x_2^2$

$x_2 = 4$

$f(x) = -x^2 + 10x - 16$

$f(4) = -(4)^2 + 40 - 16$

$= -16 + 40 - 16 = 40 - 32 = 8$

So, $Q = (4,8)$

$m = \dfrac{y_2}{x_2} = \dfrac{8}{4} = 2$

Now let's find $P$

$m = \dfrac{y_1 - 0}{10 - 0}$

$2 = \dfrac{y_1}{10}$

$y_1 = 20$

So, $P = (10, 20)$

1.21 Solution 22

Hill one: $y = 16-x^2$

Domain: $[-4,4]$

Hill two: $y = -x^2 + 12x - 32$

Domain: $[4,8]$

Let the two points touch the hills are $P$ and $Q$.

$P = (x_1,y₁)$and $Q=(x_2,y_2)$

Point $P$ is a tangent to hill one and point $Q$ is a tangent to hill two.

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$f(x) = 16 - x^2$

$f'(x) = -2x$

$g(x) = -x^2 + 12x - 32$

$g'(x) = -2x + 12$

$m = f'(x_1) = -2x_1$

$-2x_1 = \dfrac{y_2-y_1}{x_2-x_1}$

$-2x_1x_2 + 2x_1^2 = y_2 - y_1$ (Equation 1)

$m = g'(x_2) = -2x_2 + 12$

$-2x_2 + 12 = \dfrac{y_2 - y_1}{x_2 - x_1}$ (Equation 2)

$m = -2x_1$

$m = -2x_2 + 12$

$x_2 - x_1 = 6$

From 1, $y_2 - y_1 = -12x_1$

From 2, $-2x_2 + 12 = \dfrac{-12x_1}{6} = -2x_1$

$y_1 = 16 - x_1^2$

$y_2 = -x_2^2 + 12x_2 - 32$

$y_2 - y_1 = -x_2^2 + 12x_2 - 32 - 16 + x_1^2$

$= x_1^2 - x_2^2 + 12x_2 - 48$

$= (x_1 - x_2)(x_1 + x_2) + 12x_2 - 48$

$= -6(x_1 + x_2) + 12x_2 - 48$

$= 6x_2 - 6x_1 - 48$

$y_2 - y_1 = 6x_2 - 6x_1 - 48$

$= 6y(6) - 48 = 36 - 48 = -12$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-12}{6}$

$= -2$

$m = -2x_1 = -2$

$x_2 - x_1 = 6$

$x_2 - 1 = 6$

$x_2 = 7$

$y = 16-x_1^2 - 16 - 1 = 15$

$y_2 = -x_2^2 + 12x_2 - 32$

$-(7)^2 + 12.7 - 32$

$-49 + 8 - 32 = 3$

$y_2 = 3$

$y_1 = 15$

$y = mx + c$

$15 = -2.1 + c$

$c = 17$

$y = mx + c$

$y = -2x + 17$

1.22 Solution 22

Hill one: $y = 17 - x^2$

Domain: $[-\sqrt{17}, \sqrt{17}]$

Hill two: $y = -2x^2 + 26x - 80$

Domain: $[5,8]$

Let the two points of the borads touch the hills are $P$ and $Q$.

$P = (x_1, y_1)$

$Q = (x_2, y_2)$

Point $P$ is a tangent on hill one and point $Q$ is a point on the tangent of hill two.

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$f(x) = 17 - x^2$

$f'(x) = -2x$

$g(x) = -2x^2 + 26x - 80$

$g'(x) = -4x + 26$

$m = f'(x_1) = -2x_1$

$m = g'(x_2) = -4x_2 + 26$

$-2x_1 = -4x_2 + 26$

$4x_2 - 2x_1 = 26$

$2x_2 - x_1 = 13$

Also, $y_1 = 17 - x_1^2$

$y_2 = -2x_1^2 + 26x_2 - 80$

$-2x_1 = \dfrac{y_2 - y_1}{x_2 - x_1}$

$-2x_1x_2 + 2x_1^2 = y_2 - y_1$

$2x_1(x_1 - x_2) = y_2 - y_1$

$y_2 - y_1 = 2x_1(2x_1 - x_2 -x_1)$

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$m = \dfrac{-2x_1(13-x_2)}{x_2-x_1}$

$y_2 - y_1 = -2x_2^2 + 26x_2 - 80 - 17 + x_1^2$

$y_2 - y_1 = x_1^2 - 2x_2^2 + 26x_2 - 97$

$2x_1(x_1 - x_2) = x_1^2 - 2x_2^2 + 26x_2 - 97$

$2x_1^2 - 2x_1x_2 = x_1^2 - 2x_2^2 + 26x_2 - 97$

$x_1^2 - 2x_1x_2 = -2x_2^2 + 26x_2 - 97$

$x_1^2 + 2x_2^2 - 2x_1x_2 + 26x_2 + 97 = 0$

Let's try the other equation:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$-4x_2 + 26 = \dfrac{y_2 - y_1}{x_2 - x_1}$

$-4x_2 + 26(x_2 - x_1) = y_2 - y_1$

$-4x_2^2 + 4x_1x_2 + 26x_2 - 26x_1 = y_2 - y_1$

$-4x_2^2 + 4x_1x_2 + 26x_2 - 26x_1 = x_1^2 - 2x_2^2 + 26x_2 - 97$

$-2x_2^2 + 4x_1x_2 - 26x_1 = x_1^2 - 97$

$2x_1(2x_1 - x_2) -x_1^2 - 26x_1 = -97$

$2x_1(2x_1 - x_2) - x_1(x_1 + 26) = -97$

$2x_2 - x_1 = 13$

$x_2 - x_1 = 13 - x_1$

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$-2x_1 = \dfrac{y_2 - y_1}{13 - x_1}$

$-26x_1 + 2x_1^2 = y_2 - y_1$

$y_2 - y_1 = 2x_1^2 - 26x_1$

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{2x_1^2 - 26x_1}{13 - x_1}$

$-2x_1 = \dfrac{2x_1^2 - 26x_1}{13 - x_1}$

$2x_1^2 - 26x_1 = x_1^2 - 2x_2^2 + 26x_2 - 97$

$x_1^2 - 26x_1 = -2x_2^2 + 26x_2 - 97$

$x_1^2 + 2x_2^2 = 26x_1 + 26x_2 - 97$

$x_1^2 + x_2^2 + x_2^2 = 26(x_1 + x_2) - 97$

$x_1 = 2x_2 - 13$

$(2x_2 - 13)^2 + 2x_2^2 = 26(2x_2 + x_2 - 13) - 97$

$4x_2^2 + 13^2 + 2x_2^2 - 52x_2 = 26(3x_2 - 13) - 97$

$6x_2^2 + 13^2 - 52x_2 = 78x_2 - 97 - 338$

$6x_2^2 + 169 - 52x_2 = 78x_2 - 435$

$6x_2^2 - 130x_2 + 604 = 0$

$3x_2^2 - 65x_2 + 302 = 0$

Solving it we get $x_2 = \dfrac{64 \pm \sqrt{601}}{6}$

Since $x_2$ is on hill two where domain is $[5,8]$

So, $x_2 = \dfrac{65 - \sqrt{601}}{6}$

From $2x_2 - x_1 = 13$

$x_1 = 2x_2 - 13$

$x_1 = \dfrac{65 - \sqrt{601}}{3} - 13$

$x_1 = \dfrac{65 - \sqrt{601}}{3} - \dfrac{39}{3}$

$= \dfrac{26 - \sqrt{601}}{3}$

From $y_1 = 17-x_1^2$

$y_1 = 17 - \dfrac{(26 - \sqrt{601})^2}{9}$

$= \dfrac{153}{9} - \dfrac{676 + 601 - 52\sqrt{601}}{9}$

$= \dfrac{52\sqrt{601} - 1124}{9}$

$m = -2x_1 = \dfrac{2\sqrt{601}-52}{3}$

$y_1 = mx_1 + c$

$\dfrac{52\sqrt{601} - 1124}{9} - \dfrac{(2\sqrt{601} - 52)(26 - \sqrt{601})}{9} + c$

$c = \dfrac{52\sqrt{601} - 1124 - (52\sqrt{601} - 1202 - 1352 + 52\sqrt{601})}{9}$

$= \dfrac{-1124 + 1202 + 1352 - 52\sqrt{601}}{9}$

$= \dfrac{1130 - 52\sqrt{601}}{9}$

$y = mx + c$

$y = \dfrac{x(2\sqrt{601}-52)}{3} + \dfrac{1130 - 52\sqrt{601}}{9}$

$y = \dfrac{x(6\sqrt{601}-156)}{3} + \dfrac{1130 - 52\sqrt{601}}{9}$

$9y = (6\sqrt{601}-156)x + 1130 - 52\sqrt{601}$

1.23 Solution 23

$f(x) = (x^2 + 3)(2x-5)$

Method 1:

By using product rule,

$f'(x) = (x^2 + 3)(2) + (2x-5)(2x)$

$2x^2 + 6 + 4x^2 - 10x$

$= 6x^2 - 10x + 6$

Method 2:

$f(x) = (x^2 + 3)(2x -5)$

$= 2x^2 - 5x^2 + 6x - 15$

$f'(x) = 6x - 10x + 6$

1.24 Solution 24

$a^4 - b^4$

1.24.1 Solution a

$(a-b)? = a^4 - b^4$

$a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$

$= (a^2 + b^2)(a+b)(a-b)$

$= (a^3 + a^2b + ab^2 + b^3)(a-b)$

1.24.2 Solution b

$f(x) = \sqrt[4]{x}$

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

$= \lim_{h \to 0} \dfrac{\sqrt[4]{x+h} - \sqrt[4]{x}}{h}$

From $(a)$, let $a = \sqrt[4]{x+h}$ and $b = \sqrt[4]{x}$

$(x+h)-h = ((\sqrt[4]{x+h})^3 + (\sqrt[4]{x+h})^2\sqrt[4]{x} + \sqrt[4]{x+h}\sqrt{x} + x^{3/4})(\sqrt[4]{x+h} - \sqrt[4]{x})$

$f'(x) = \lim_{h \to 0} \dfrac{\sqrt[4]{x+h} - \sqrt[4]{x}}{h} * \dfrac{((x+h)^{3/4} + (x+h)^{1/2}\sqrt[4]{x} + \sqrt[4]{x+h} + \sqrt{x} + x^{3/4})}{((x+h)^{3/4} + (x+h)^{1/2}\sqrt[4]{x} + \sqrt[4]{x+h} + \sqrt{x} + x^{3/4})}$

$\lim_{h \to 0} \dfrac{h}{h(((x+h)^{3/4} + (x+h)^{1/2}\sqrt[4]{x} + \sqrt[4]{x+h} + \sqrt{x} + x^{3/4}))}$

$\lim_{h \to 0} \dfrac{1}{((x+h)^{3/4} + (x+h)^{1/2}\sqrt[4]{x} + \sqrt[4]{x+h} + \sqrt{x} + x^{3/4})}$

Let's take the limit of the denomiator in the last fraction, we get:

$\lim_{h \to 0} ((x+h)^{3/4} + (x+h)^{1/2}\sqrt[4]{x} + \sqrt[4]{x+h} + \sqrt{x} + x^{3/4})$

$= x^{3/4} + x^{1/2}x^{1/4} + x^{1/4}x^{1/2} + x^{3/4}$

$= 2x^{3/4} + 2x^{1/2}x^{1/4}$

$= 2x^{3/4} + 2x^{3/4}$

$= 4x^{3/4}$

Thus as long as $x \neq 0$, we have

$f'(x) = \dfrac{1}{4(\sqrt[4]{x})^2}$

1.24.3 Solution c

$n \geq 2$

$(a^n - b^n) = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + .... + ab^{n-2} + b^{n-2})$

$a = \sqrt[n]{x+h}$

$b = \sqrt[n]{x}$

$f'(x) = \lim_{h \to 0} \dfrac{\sqrt[n]{x+h} - \sqrt[n]{x}}{h} * \dfrac{((\sqrt[n]{x+h})^{n-1} + (\sqrt[n]{x+h})^{n-2}\sqrt[n]{x} + .... \sqrt[n]{x+h}(\sqrt[n]{x})^{n-2}) + (\sqrt[n]{x})^{n-1}}{((\sqrt[n]{x+h})^{n-1} + (\sqrt[n]{x+h})^{n-2}\sqrt[n]{x} + .... \sqrt[n]{x+h}(\sqrt[n]{x})^{n-2}) + (\sqrt[n]{x})^{n-1}}$

$f'(x) = \lim_{h \to 0} \dfrac{h}{h(((\sqrt[n]{x+h})^{n-1} + (\sqrt[n]{x+h})^{n-2}\sqrt[n]{x} + .... \sqrt[n]{x+h}(\sqrt[n]{x})^{n-2}) + (\sqrt[n]{x})^{n-1})}$

$f'(x) = \lim_{h \to 0} \dfrac{1}{((\sqrt[n]{x+h})^{n-1} + (\sqrt[n]{x+h})^{n-2}\sqrt[n]{x} + .... \sqrt[n]{x+h}(\sqrt[n]{x})^{n-2}) + (\sqrt[n]{x})^{n-1}}$

Let's take the limit of the denominator in the last fraction, we get:

$\lim_{h \to 0} ((\sqrt[n]{x+h})^{n-1} + (\sqrt[n]{x+h})^{n-2}\sqrt[n]{x} + .... \sqrt[n]{x+h}(\sqrt[n]{x})^{n-2}) + (\sqrt[n]{x})^{n-1}$

Let $\sqrt[n]{x} = c$

$= c^{n-1} + c^{n-2}.c + c^{n-3}.c^2 + ... + c.c^{n-2} + c^{n-1}$

$= c^{n-1} + c^{n-1} + c^{n-1} + ... c^{n-1} + c^{n-1}$

$= nc^{n-1}$

$= n(\sqrt[n]{x})^{n-1}$

$f'(x) = \dfrac{1}{n(\sqrt[n]{x})^{n-1}} = \dfrac{(\sqrt[n]{x})^{1-n}}{n}$

$= \dfrac{1}{n}x^{1/n - 1}$

1.25 Solution 25

1.25.1 Solution a

$\dfrac{d}{dx} \sec x = \sec x \tan x$

$\dfrac{d}{dx} \sec x = \dfrac{d}{dx} (\dfrac{1}{\cos x})$

Using the quotient rule,

$\dfrac{d}{dx}(\dfrac{1}{\cos x}) = \dfrac{\cos x 0 - 1(-\sin x)}{(\cos x)^2}$

$= 0 + \dfrac{\sin x}{(\cos x)^2} = \dfrac{\tan x}{\cos x} = \sec x \tan x$

1.25.2 Solution b

$\dfrac{d}{dx} \cot x = -\csc^2 x$

$\dfrac{d}{dx} \cot x = \dfrac{d}{dx} \dfrac{\cos x}{\sin x}$

Using the quotient rule,

$\dfrac{d}{dt} \dfrac{\cos x}{\sin x} = \dfrac{\sin x (-\sin x) - \cos x(\cos x)}{(\sin x)^2}$

$= \dfrac{-\sin^2x - \cos^2 x}{\sin^2 x}$

$= \dfrac{-1(\sin^2 x + \cos^2 x)}{\sin^2 x}$

$= \dfrac{-1}{\sin^2 x} = -\csc^2 x$

1.25.3 Solution c

$\dfrac{d}{dx}(\csc x) = -\csc x \cot x$

$\dfrac{d}{dx} (\csc x) = \dfrac{d}{dx} (\dfrac{1}{\sin x})$

Using the quotient rule,

$\dfrac{d}{dx} (dfrac{1}{\sin x}) = \dfrac{\sin x.0 - 1\cos x}{(\sin x)^2}$

$= \dfrac{-\cos x}{\sin^2 x}$

$= \dfrac{-\cot x}{\sin x}$

$= -\csc x \cot x$

1.26 Solution 26

$f$ is differentiable on $(-\infty, \infty)$

1.26.1 Solution a

$\dfrac{d}{dx} ((f(x))^2) = 2f(x)f'(x)$

Let $g(x) = (f(x))^2$

We need to prove $g'(x) = 2f(x)f'(x)$

$g'(x) = \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}$

$= \lim_{h \to 0} \dfrac{f(x+h)^2 - f(x)^2}{h}$

$= \lim_{h \to 0} \dfrac{(f(x+h)-f(x))(f(x+h)+f(x))}{h}$

$= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} * \lim_{h \to 0} \dfrac{f(x+h)+f(x)}{1}$

$= f'(x)(f(x) + f(x))$

$= 2f(x)f'(x)$

1.26.2 Solution b

$\dfrac{d}{dx} ((f(x))^3)$

Let $g(x) = (f(x))^3$

$g'(x) = \lim_{h \to 0} \dfrac{g(x+h)-g(x)}{h}$

$= \lim_{h \to 0} \dfrac{f(x+h)^3 - f(x)^3}{h}$

We know that $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

$= \lim_{h \to 0} \dfrac{(f(x+h) - f(x))(f(x+h)^2 + f(x+h)f(x) + f(x)^2)}{h}$

$= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} * \lim_{h \to 0} f(x+h)^2 + f(x+h)f(x) + f(x)^2$

$= f'(x)(f(x)^2 + f(x)f(x) + f(x)^2)$

$= 3f(x)^2f'(x)$

1.26.3 Solution c

Conjecture: $\dfrac{d}{dx} ((f(x))^n) = nf(x)^{n-1}f'(x)$

We will prove using mathematical induction.

Base case: $n = 1$

$\dfrac{d}{dx} f(x) = 1(f(x))^0f'(x) = f'(x)$

Induction step: Suppose $n$ is a positive integer and $\dfrac{d}{dx} ((f(x))^n) = nf(x)^{n-1}f'(x)$

$\dfrac{d}{dx} (f(x))^{n+1} = \dfrac{d}{dx} f(x)f(x)^n$

Using the product rule,

$= f(x)\dfrac{d}{dx}(f(x))^n + f(x)^n \dfrac{d}{dx} f(x)$

$= f(x)nf(x)^{n-1}f'(x) + f(x)^nf'(x)$

$= f(x)^n . n . f'(x) + f(x)^nf'(x)$

$= f(x)^nf'(x)(n+1)$

This completes the mathematical induction proof.

1.27 Solution 27

$\dfrac{d}{dx} f(x)g(x)h(x)$

Using the product rule,

$= f(x)\dfrac{d}{dx}g(x)h(x) + g(x)h(x)f'(x)$

$= f(x)(g(x)h'(x) + h(x)g'(x)) + g(x)h(x)f'(x)$

$= f(x)g(x)h'(x) + f(x)h(x)g'(x) + g(x)h(x)f'(x)$