(1.) Determine $f'(x)$ from second principles if it is given $f(x) = x^2 - 5$
Power Rule and Difference Rule
$
f(x) = x^2 - 5 \\[3ex]
f(x) = x^2 - 5x^0 \\[3ex]
f'(x) = 2 * x^{2 - 1} - 0 * 5x^{0 - 1} \\[3ex]
= 2 * x^1 - 0 \\[3ex]
= 2x
$
(2.) Differentiate, with respect to
x, $y = 3x^2 + 4x - 1$, from second principles.
Power Rule and Sum/Difference Rule
$
y = 3x^2 + 4x - 1 \\[3ex]
y = 3x^2 + 4x^1 - 1x^0 \\[3ex]
\dfrac{dy}{dx} = 2 * 3 x^{2 - 1} + 1 * 4x^{1 - 1} - 0 * x^{0 - 1} \\[3ex]
= 2 * 3x + 1 * 4x^0 - 0 \\[3ex]
= 6x + 4
$
(3.) Differentiate from second principles, with respect to
x,
$y = 3x^2 + 2x - 1$
Power Rule and Sum/Difference Rule
$
y = 3x^2 + 2x - 1 \\[3ex]
y = 3x^2 + 2x^1 - 1x^0 \\[3ex]
\dfrac{dy}{dx} = 2 * 3 x^{2 - 1} + 1 * 2x^{1 - 1} - 0 * x^{0 - 1} \\[3ex]
= 2 * 3x + 1 * 2x^0 - 0 \\[3ex]
= 6x + 2
$
(4.) Determine the derivative of $x - 2x^3$ by rules.
Power Rule and Difference Rule
$
f(x) = x - 2x^3 \\[3ex]
f(x) = x^1 - 2x^3 \\[3ex]
f'(x) = 1 * x^{1 - 1} - 3 * 2x^{3 - 1} \\[3ex]
= x^0 - 3 * 2x^2 \\[3ex]
= 1 - 6x^2
$
(5.) Differentiate $y = \dfrac{1}{x}$
wrt $x$ by rules.
Power Rule
$
y = \dfrac{1}{x} = x^{-1} \\[5ex]
\dfrac{dy}{dx} = -1 * x^{-1 - 1} \\[5ex]
= -x^{-2} \\[5ex]
= -\dfrac{1}{x^2}
$
(6.) Find the derivative of $y = \sqrt{x}$ by second principle (by rules)
Power Rule
$
y = \sqrt{x} \\[3ex]
y = x^{\dfrac{1}{2}} \\[5ex]
\dfrac{dy}{dx} = \dfrac{1}{2} * x^{\dfrac{1}{2} - 1} \\[5ex]
= \dfrac{1}{2} * x^{\dfrac{1}{2} - \dfrac{2}{2}} \\[5ex]
= \dfrac{1}{2} * x^-{\dfrac{1}{2}} \\[5ex]
= \dfrac{1}{2} * \dfrac{1}{x^{\dfrac{1}{2}}} \\[7ex]
= \dfrac{1}{2} * \dfrac{1}{\sqrt{x}} \\[7ex]
= \dfrac{1}{2\sqrt{x}}
$
(7.)
USSCE:Advance Mathematics Paper 1 Find $\dfrac{dy}{dx}$ is $y = \dfrac{1}{8}$
$
y = \dfrac{1}{8} \\[5ex]
\dfrac{dy}{dx} = 0 \\[3ex]
$
The derivative of any constant is zero
Alternative Explanation
Power Rule
$
y = \dfrac{1}{8} \\[5ex]
y = \dfrac{1}{8} * 1 \\[5ex]
y = \dfrac{1}{8}x^0 \\[5ex]
\dfrac{dy}{dx} = 0 * \dfrac{1}{8}* x^{0 - 1} \\[5ex]
= 0
$
(8.)
USSCE: Advance Mathematics Paper 1 The first derivative of $y = x^3 + 2x^2 + 3x - 4$ is
$
A.\;\; 3x^3 + 2x + 3 \\[3ex]
B.\;\; 3x^2 + 2x + 3 \\[3ex]
C.\;\; 3x^2 + 4x + 3 \\[3ex]
D.\;\; 3x^2 + 4^2x + 3 \\[3ex]
$
Power Rule
$
y = x^3 + 2x^2 + 3x - 4 \\[3ex]
y' = 3x^2 + 4x + 3
$
(9.)
NSC Determine $\dfrac{dy}{dx}$ if:
$
(9.1)\:\:y = 3x^3 + 6x^2 + x - 4 \\[3ex]
(9.2)\:\: yx - y = 2x^2 - 2x;\:\:x \ne 1 \\[3ex]
$
$
(9.1) \\[3ex]
y = 3x^3 + 6x^2 + x - 4 \\[3ex]
\dfrac{dy}{dx} = 9x^2 + 12x + 1 \\[5ex]
(9.2) \\[3ex]
yx - y = 2x^2 - 2x \\[3ex]
\text{Factor by GCF} \\[3ex]
y(x - 1) = 2x(x - 1) \\[3ex]
y = \dfrac{2x(x - 1)}{x - 1} \\[5ex]
y = 2x \\[3ex]
\dfrac{dy}{dx} = 2
$
(10.)
JAMB If $y = x^2 - \dfrac{1}{x}$, find $\dfrac{dy}{dx}$
$
A.\:\: 2x + \dfrac{1}{x^2} \\[5ex]
B.\:\: 2x + x^2 \\[3ex]
C.\:\: 2x - \dfrac{1}{x^2} \\[5ex]
D.\:\: 2x - x^2 \\[3ex]
$
Power Rule
$
y = x^2 - \dfrac{1}{x} \\[5ex]
y = x^2 - x^{-1} \\[3ex]
\dfrac{dy}{dx} = 2 * x^{2 - 1} - -1 * x^{-1 - 1} \\[5ex]
= 2 * x^1 + 1 * x^{-2} \\[5ex]
= 2x + x^{-2} \\[5ex]
= 2x + \dfrac{1}{x^2}
$
(11.)
JAMB If $y = x\sin x$, find $\dfrac{dy}{dx}$ when $x = \dfrac{\pi}{2}$
$
A.\:\: -\dfrac{\pi}{2} \\[5ex]
B.\:\: -1 \\[3ex]
C.\:\: 1 \\[3ex]
D.\:\: \dfrac{\pi}{2} \\[5ex]
$
Product Rule, Power Rule
$
y = x\sin x \\[3ex]
u = x \\[3ex]
u' = 1 \\[3ex]
v = \sin x \\[3ex]
v' = \cos x \\[3ex]
y' = uv' + vu' ...Product\:\:Rule \\[3ex]
y' = x(\cos x) + \sin x(1) \\[3ex]
y' = x\cos x + \sin x \\[3ex]
x = \dfrac{\pi}{2} = \dfrac{180^\circ}{2} = 90^\circ \\[5ex]
y' = \dfrac{\pi}{2}\cos90^\circ + \sin90^\circ \\[5ex]
\cos 90^\circ = 0 \\[3ex]
\sin 90^\circ = 1 \\[3ex]
\rightarrow y' = \dfrac{\pi}{2} * 0 + 1 \\[5ex]
y' = 0 + 1 \\[3ex]
y' = 1
$
(12.)
JAMB Differentiate $(2x + 5)^2(x - 4)$ with respect to $x$
$
A.\:\: 4(2x + 5)(x - 4) \\[3ex]
B.\:\: 4(2x + 5)(4x - 3) \\[3ex]
C.\:\: (2x + 5)(2x - 13) \\[3ex]
D.\:\: (2x + 5)(6x - 11) \\[3ex]
$
Product Rule, Sum/Difference Rule, Power Rule
$
y = (2x + 5)^2(x - 4) \\[3ex]
u = (2x + 5)^2 = (2x + 5)(2x + 5) = 4x^2 + 10x + 10x + 25 = 4x^2 + 20x + 25 \\[3ex]
\dfrac{du}{dx} = 8x + 20 \\[5ex]
v = x - 4 \\[3ex]
\dfrac{dv}{dx} = 1 \\[5ex]
\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}...Product\:\:Rule \\[5ex]
= (4x^2 + 20x + 25)(1) + (x - 4)(8x + 20) \\[5ex]
= 4x^2 + 20x + 25 + 8x^2 + 20x - 32x - 80 \\[5ex]
= 12x^2 + 8x - 55 \\[3ex]
$
This is JAMB exam. You need to solve this within a minute without a calculator.
A quick review at the options shows that option $D.$ is the answer
$
\dfrac{dy}{dx} = (2x + 5)(6x - 11)
$
(13.)
ATAR Use the quotient rule to show that
$\dfrac{d}{dx} \tan(x) = \dfrac{1}{\cos^2(x)}$
Quotient Rule
$
f(x) = \tan x \\[3ex]
= \dfrac{\sin x}{\cos x} ...\text{Quotient Identity} \\[5ex]
For\;\;f(x) = \dfrac{u}{v} \\[5ex]
u = \sin x \\[3ex]
u' = \cos x \\[3ex]
v = \cos x \\[3ex]
v' = -\sin x \\[3ex]
f'(x) = \dfrac{vu' - uv'}{v^2} \\[5ex]
\underline{Numerator} \\[3ex]
vu' - uv' \\[3ex]
= (\cos x)(\cos x) - (\sin x)(-\sin x) \\[3ex]
= \cos^2 x + \sin^2 x \\[3ex]
= 1...Pythagorean\;\;Identity \\[3ex]
\underline{Denominator} \\[3ex]
v^2 = \cos^2 x \\[3ex]
\implies \\[3ex]
f'(x) = \dfrac{1}{\cos^2 x} \\[5ex]
\therefore \dfrac{d}{dx} \tan(x) = \dfrac{1}{\cos^2(x)}
$
(14.)
NZQA Differentiate $y = \left(3x - x^2\right)^5$
You do not need to simplify your answer.
Function of a Function Rule, Power Rule
$
y = \left(3x - x^2\right)^5 \\[3ex]
Let\;\; p = 3x - x^2 \;\;\; means\;\;that\;\; y = p^5 \\[3ex]
\dfrac{dp}{dx} = 3 - 2x \\[5ex]
\dfrac{dy}{dp} = 5p^4 \\[3ex]
\dfrac{dy}{dx} = \dfrac{dy}{dp} * \dfrac{dp}{dx} ...Chain\;\;Rule \\[5ex]
= 5p^4(3 - 2x) \\[3ex]
= 5(3x - x^2)^4(3 - 2x) \\[3ex]
= 5(3 - 2x)(3x - x^2)^4
$
(15.)
NZQA Differentitate $y = \dfrac{\tan x}{x^3}$
You do not need to simplify your answer.
Quotient Rule, Power Rule
$
f(x) = \dfrac{\tan x}{x^3} \\[5ex]
For\;\;f(x) = \dfrac{u}{v} \\[5ex]
u = \tan x \\[3ex]
u' = \sec^2 x \\[3ex]
v = x^3 \\[3ex]
v' = 3x^2 \\[3ex]
f'(x) = \dfrac{vu' - uv'}{v^2} \\[5ex]
\underline{Numerator} \\[3ex]
vu' - uv' \\[3ex]
= (x^3)(\sec^2 x) - (\tan x)(3x^2) \\[3ex]
= x^3\sec^2 x - 3x^2\tan x \\[3ex]
\underline{Denominator} \\[3ex]
v^2 = (x^3)^2 = x^6 \\[3ex]
\implies \\[3ex]
f'(x) \\[3ex]
= \dfrac{x^3\sec^2 x - 3x^2\tan x}{x^6} \\[5ex]
= \dfrac{x^2(x\sec^2 x - 3\tan x)}{x^6} \\[5ex]
= \dfrac{x\sec^2 x - 3\tan x}{x^4}
$
(21.) Differentiate $\ln|\ln|\ln x||\:\:wrt\:\:x$
Chain Rule, Standard Derivatives
$
y = \ln|\ln|\ln x|| \\[3ex]
Let: \\[3ex]
p = \ln x \:\:\:\: \dfrac{dp}{dx} = \dfrac{1}{x} \\[5ex]
q = |p| \:\:\:\: \dfrac{dq}{dp} = \dfrac{|p|}{p} \\[5ex]
r = \ln q \:\:\:\: \dfrac{dr}{dq} = \dfrac{1}{q} \\[5ex]
s = |r| \:\:\:\: \dfrac{ds}{dr} = \dfrac{|r|}{r} \\[5ex]
y = \ln s \:\:\:\: \dfrac{dy}{ds} = \dfrac{1}{s} \\[5ex]
\dfrac{dy}{dx} = \dfrac{dp}{dx} * \dfrac{dq}{dp} * \dfrac{dr}{dq} * \dfrac{ds}{dr} *
\dfrac{dy}{ds} \\[5ex]
\dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{|p|}{p} * \dfrac{1}{q} * \dfrac{|r|}{r} * \dfrac{1}{s}
\\[5ex]
Substitute\:\:back \\[3ex]
\dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{|\ln x|}{\ln x} * \dfrac{1}{|\ln x|} * \dfrac{|\ln
q|}{\ln q} * \dfrac{1}{|\ln
q|} \\[5ex]
\dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln q} \\[5ex]
= \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln |p|} \\[5ex]
= \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln |\ln x|} \\[5ex]
= \dfrac{1 * 1 * 1}{x * \ln x * \ln|\ln x|} \\[5ex]
= \dfrac{1}{x\ln x\ln|\ln x|}
$
(22.) Differentiate these functions with respect to (
wrt)
x
$
(a.)\;\; f(x) = \left(\dfrac{x + 2}{x + 4}\right)^9 \\[5ex]
(b.)\;\; y = x^5\sin^2 x\cos 4x \\[3ex]
(c.)\;\; y = xy - 2x^2y^2 - 5x \\[3ex]
(d.)\;\; y = (x^2 - 1)(3 - 2x)^2 \\[3ex]
$
$
(a.) \\[3ex]
f(x) = \left(\dfrac{x + 2}{x + 4}\right)^9 \\[5ex]
y = \left(\dfrac{x + 2}{x + 4}\right)^9 \\[5ex]
\text{Let }p = \dfrac{x + 2}{x - 4} \hspace{3em} \implies y = p^9 \\[5ex]
........................................................... \\[3ex]
p = \dfrac{x + 2}{x - 4} = \dfrac{u}{v} \\[5ex]
\text{Let }u = x + 2 \hspace{5em} v = x - 4 \\[3ex]
u' = \dfrac{du}{dx} = 1 \hspace{6em} v' = \dfrac{dv}{dx} = 1 ...\text{Power Rule} \\[5ex]
\dfrac{dp}{dx} = \dfrac{vu' - uv'}{v^2}...\text{Quotient Rule} \\[5ex]
= \dfrac{(x - 4)(1) - (x + 2)(1)}{(x - 4)^2} \\[5ex]
= \dfrac{x - 4 - x - 2}{(x - 4)^2} \\[5ex]
= -\dfrac{6}{(x - 4)^2} \\[5ex]
........................................................... \\[3ex]
y = p^9 \\[3ex]
\dfrac{dy}{dp} = 9p^8 ...\text{Power Rule} \\[3ex]
........................................................... \\[3ex]
\dfrac{dy}{dx} = \dfrac{dy}{dp} * \dfrac{dp}{dx} ...\text{Function of a Function Rule} \\[5ex]
= 9p^8 * -\dfrac{6}{(x - 4)^2} \\[5ex]
= -54p^8 \div (x - 4)^2 \\[3ex]
= -54\left(\dfrac{x + 2}{x - 4}\right)^8 * \dfrac{1}{(x - 4)^2} \\[5ex]
= -\dfrac{54(x + 2)^8}{(x - 4)^8 * (x - 4)^2} \\[5ex]
= -\dfrac{54(x + 2)^8}{(x - 4)^{10}}
$
$
(b.) \\[3ex]
y = x^5\sin^2 x\cos 4x \\[3ex]
y = x^5 * \sin^2 x * \cos 4x \\[3ex]
y = u * v * w \\[3ex]
Let\;\;u = x^5 \hspace{3em} v = \sin^2 x \hspace{3em} w = \cos 4x \\[3ex]
.................................................................... \\[3ex]
u = x^5 \\[3ex]
u' = 5x^4...\text{Power Rule} \\[5ex]
v = \sin^2 x \\[3ex]
Let\;\; m = \sin x \hspace{2em}\implies\hspace{2em} v = m^2 \\[3ex]
\dfrac{dm}{dx} = \cos x \hspace{5em} \dfrac{dv}{dm} = 2m \\[5ex]
\dfrac{dv}{dx} = \dfrac{dv}{dm} * \dfrac{dm}{dx} ...\text{Function of a Function Rule} \\[5ex]
v' = 2m * \cos x \\[3ex]
v' = 2\sin x \cos x \\[5ex]
w = \cos 4x \\[3ex]
Let\;\; n = 4x \hspace{2em}\implies\hspace{2em} w = \cos n \\[3ex]
\dfrac{dn}{dx} = 4 \hspace{5em} \dfrac{dw}{dn} = -\sin n \\[5ex]
\dfrac{dw}{dx} = \dfrac{dw}{dn} * \dfrac{dn}{dx} ...\text{Function of a Function Rule} \\[5ex]
w' = -\sin n * 4 \\[3ex]
w' = -4\sin 4x \\[3ex]
.................................................................... \\[3ex]
y' = u'vw + uv'w + uvw' ...\text{Product Rule} \\[3ex]
= 5x^4(\sin^2 x)(\cos 4x) + x^5(2\sin x \cos x)(\cos 4x) + x^5(\sin^2 x)(-4\sin 4x) \\[3ex]
= 5x^4\sin^2 x\cos 4x + 2x^5\sin x\cos x\cos 4x - 4x^5\sin^2x\sin 4x \\[3ex]
= x^4\sin x[5\sin x\cos 4x + 2x\cos x\cos 4x - 4x\sin x\sin 4x] \\[3ex]
= x^4\sin x[\cos 4x(5\sin x + 2x\cos x) - 4x\sin x\sin 4x]
$
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