Question 1
Use the quotient rule: when $y=\frac uv$, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{u'v-v'u}{v^2}$ to differentiate the following:
a) $y=\frac{x+1}{x+2}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{(x+2)}^2}$
Question ID: 50040010010
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b) $y=\frac{2x+1}{x+1}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{(x+1)}^2}$
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c) $y=\frac{2x-3}{4x+5}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2}{{(4x+5)}^2}$
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d) $y=\frac{4-x}{6x-10}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{14}{{(6x+10)}^2}$
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e) $y=\frac{5x}{20x+11}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{55}{{(20x+11)}^2}$
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f) $y=\frac{1-2x}{5x+6}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{17}{{(5x+6)}^2}$
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g) $y=\frac{2x+5}{7-3x}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{29}{{(7-3x)}^2}$
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h) $y=-\frac{33-2x}{5+2x}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{76}{{(5+2x)}^2}$
Question ID: 50040010080
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i) $y=\frac{ax+b}{a+bx}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{a^2-b^2}{{(a+bx)}^2}$
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Question 2
Use the quotient rule to differentiate the following:
a) $y=\frac{x^2+1}{x^2+2}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2x}{{(x^2+2)}^2}$
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b) $y=\frac{x^3-5}{x^2+6}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^4+18x^2+10x}{{(x^2+6)}^2}$
Question ID: 50040020020
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c) $y=\frac{1-x^4}{2x+3}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{6x^4+12x^3+2}{{(2x+3)}^2}$
Question ID: 50040020030
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d) $y=\frac{3x^2-9}{2x^2+1}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{42x}{{(2x^2+1)}^2}$
Question ID: 50040020040
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e) $y=\frac{6x^2+x}{2x-4}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{12x^2-48x-4}{{(2x-4)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x^2-12x-1}{{(x-2)}^2}$
Question ID: 50040020050
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f) $y=\frac{4x^3-6x+1}{2x-3}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{16x^3-36x^2+16}{{(2x-3)}^2}$
Question ID: 50040020060
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g) $y=\frac{4x^2+6x+3}{x^2-5x+4}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{39+26x-26x^2}{{(x^2-5x+4)}^2}$
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Question 3
Use the quotient rule to differentiate the following:
a) $y=\frac{\sqrt{x}}{x+2}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^{-\frac12}-\frac12x^{\frac12}}{{(x+2)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^{-\frac12}(x-\frac x2)}{{(x+2)}^2}$
Question ID: 50040030010
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b) $y=\frac{\sqrt{x}+1}{\sqrt{x}-1}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x^{-\frac12}}{{(\sqrt{x}-1)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{x}{(\sqrt{x}-1)}^2}$
Question ID: 50040030020
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c) $y=\frac{\sqrt{x+5}}{2x+3}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{2x+3}{2\sqrt{x+5}}-2\sqrt{x+5}}{{(2x+3)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{23-2x}{2\sqrt{x+5}{(2x+5)}^2}$
Question ID: 50040030030
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d) $y=\frac{\sqrt{x}-\sqrt3}{\sqrt{x-3}}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\sqrt{x-3}}{2\sqrt{x}}-\frac{\sqrt{x}-\sqrt{3}}{2\sqrt{x-3}}}{x-3}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sqrt{3}\sqrt{x}-3}{2\sqrt{{(x-3)}^3}\sqrt{x}}$
Question ID: 50040030040
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e) $y=\frac{x{(x+2)}^2}{4x^3-8}$
a
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4(x+2)(3x^4+2x^3-4x-2)}{{(4x^3-8)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x+2)(2x^3+3x+2)}{2{(x^3-2)}^2}$
Question ID: 50040030050
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Question 1
Use the quotient rule: when $y=\frac uv$, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{u'v-v'u}{v^2}$ to differentiate the following:
a) $y=\frac{x+1}{x+2}$
b) $y=\frac{2x+1}{x+1}$
c) $y=\frac{2x-3}{4x+5}$
d) $y=\frac{4-x}{6x-10}$
e) $y=\frac{5x}{20x+11}$
f) $y=\frac{1-2x}{5x+6}$
g) $y=\frac{2x+5}{7-3x}$
h) $y=-\frac{33-2x}{5+2x}$
i) $y=\frac{ax+b}{a+bx}$
Question 2
Use the quotient rule to differentiate the following:
a) $y=\frac{x^2+1}{x^2+2}$
b) $y=\frac{x^3-5}{x^2+6}$
c) $y=\frac{1-x^4}{2x+3}$
d) $y=\frac{3x^2-9}{2x^2+1}$
e) $y=\frac{6x^2+x}{2x-4}$
f) $y=\frac{4x^3-6x+1}{2x-3}$
g) $y=\frac{4x^2+6x+3}{x^2-5x+4}$
Question 3
Use the quotient rule to differentiate the following:
a) $y=\frac{\sqrt{x}}{x+2}$
b) $y=\frac{\sqrt{x}+1}{\sqrt{x}-1}$
c) $y=\frac{\sqrt{x+5}}{2x+3}$
d) $y=\frac{\sqrt{x}-\sqrt3}{\sqrt{x-3}}$
e) $y=\frac{x{(x+2)}^2}{4x^3-8}$
Answers
Question 1
a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{(x+2)}^2}$
b) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{(x+1)}^2}$
c) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2}{{(4x+5)}^2}$
d) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{14}{{(6x+10)}^2}$
e) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{55}{{(20x+11)}^2}$
f) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{17}{{(5x+6)}^2}$
g) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{29}{{(7-3x)}^2}$
h) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{76}{{(5+2x)}^2}$
i) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{a^2-b^2}{{(a+bx)}^2}$
Question 2
a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2x}{{(x^2+2)}^2}$
b) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^4+18x^2+10x}{{(x^2+6)}^2}$
c) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{6x^4+12x^3+2}{{(2x+3)}^2}$
d) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{42x}{{(2x^2+1)}^2}$
e) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{12x^2-48x-4}{{(2x-4)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x^2-12x-1}{{(x-2)}^2}$
f) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{16x^3-36x^2+16}{{(2x-3)}^2}$
g) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{39+26x-26x^2}{{(x^2-5x+4)}^2}$
Question 3
a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^{-\frac12}-\frac12x^{\frac12}}{{(x+2)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^{-\frac12}(x-\frac x2)}{{(x+2)}^2}$
b) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x^{-\frac12}}{{(\sqrt{x}-1)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{x}{(\sqrt{x}-1)}^2}$
c) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{2x+3}{2\sqrt{x+5}}-2\sqrt{x+5}}{{(2x+3)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{23-2x}{2\sqrt{x+5}{(2x+5)}^2}$
d) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\sqrt{x-3}}{2\sqrt{x}}-\frac{\sqrt{x}-\sqrt{3}}{2\sqrt{x-3}}}{x-3}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sqrt{3}\sqrt{x}-3}{2\sqrt{{(x-3)}^3}\sqrt{x}}$
e) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4(x+2)(3x^4+2x^3-4x-2)}{{(4x^3-8)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x+2)(2x^3+3x+2)}{2{(x^3-2)}^2}$
Question 1
Use the quotient rule: when $y=\frac uv$, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{u'v-v'u}{v^2}$ to differentiate the following:
a) $y=\frac{x+1}{x+2}$
b) $y=\frac{2x+1}{x+1}$
c) $y=\frac{2x-3}{4x+5}$
d) $y=\frac{4-x}{6x-10}$
e) $y=\frac{5x}{20x+11}$
f) $y=\frac{1-2x}{5x+6}$
g) $y=\frac{2x+5}{7-3x}$
h) $y=-\frac{33-2x}{5+2x}$
i) $y=\frac{ax+b}{a+bx}$
Question 2
Use the quotient rule to differentiate the following:
a) $y=\frac{x^2+1}{x^2+2}$
b) $y=\frac{x^3-5}{x^2+6}$
c) $y=\frac{1-x^4}{2x+3}$
d) $y=\frac{3x^2-9}{2x^2+1}$
e) $y=\frac{6x^2+x}{2x-4}$
f) $y=\frac{4x^3-6x+1}{2x-3}$
g) $y=\frac{4x^2+6x+3}{x^2-5x+4}$
Question 3
Use the quotient rule to differentiate the following:
a) $y=\frac{\sqrt{x}}{x+2}$
b) $y=\frac{\sqrt{x}+1}{\sqrt{x}-1}$
c) $y=\frac{\sqrt{x+5}}{2x+3}$
d) $y=\frac{\sqrt{x}-\sqrt3}{\sqrt{x-3}}$
e) $y=\frac{x{(x+2)}^2}{4x^3-8}$
Answers
Question 1
a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{(x+2)}^2}$
b) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{(x+1)}^2}$
c) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2}{{(4x+5)}^2}$
d) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{14}{{(6x+10)}^2}$
e) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{55}{{(20x+11)}^2}$
f) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{17}{{(5x+6)}^2}$
g) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{29}{{(7-3x)}^2}$
h) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{76}{{(5+2x)}^2}$
i) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{a^2-b^2}{{(a+bx)}^2}$
Question 2
a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2x}{{(x^2+2)}^2}$
b) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^4+18x^2+10x}{{(x^2+6)}^2}$
c) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{6x^4+12x^3+2}{{(2x+3)}^2}$
d) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{42x}{{(2x^2+1)}^2}$
e) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{12x^2-48x-4}{{(2x-4)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{3x^2-12x-1}{{(x-2)}^2}$
f) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{16x^3-36x^2+16}{{(2x-3)}^2}$
g) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{39+26x-26x^2}{{(x^2-5x+4)}^2}$
Question 3
a) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^{-\frac12}-\frac12x^{\frac12}}{{(x+2)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^{-\frac12}(x-\frac x2)}{{(x+2)}^2}$
b) $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x^{-\frac12}}{{(\sqrt{x}-1)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{x}{(\sqrt{x}-1)}^2}$
c) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{2x+3}{2\sqrt{x+5}}-2\sqrt{x+5}}{{(2x+3)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{23-2x}{2\sqrt{x+5}{(2x+5)}^2}$
d) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\sqrt{x-3}}{2\sqrt{x}}-\frac{\sqrt{x}-\sqrt{3}}{2\sqrt{x-3}}}{x-3}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sqrt{3}\sqrt{x}-3}{2\sqrt{{(x-3)}^3}\sqrt{x}}$
e) $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4(x+2)(3x^4+2x^3-4x-2)}{{(4x^3-8)}^2}$
or
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{(x+2)(2x^3+3x+2)}{2{(x^3-2)}^2}$