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User:Michael Hardy
d
y
d
x
=
d
d
x
(
(
2
x
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1
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3
(
3
x
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5
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=
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2
x
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1
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3
⋅
d
d
x
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3
x
−
1
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5
+
(
3
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5
d
d
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2
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1
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3
=
(
2
x
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3
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5
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3
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4
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3
+
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3
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3
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3
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2
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2
⋅
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15
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2
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3
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4
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6
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3
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3
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3
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3
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2
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15
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6
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3
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15
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2
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10
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6
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3
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3
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15
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x
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5
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85
15
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x
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5
−
85
15
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=
90
⋅
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3
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2
⋅
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3
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3
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x
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15
)
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}={}&{\frac {d}{dx}}{\Big (}(2x+1)^{3}(3x-1)^{5}{\Big )}=(2x+1)^{3}\cdot {\frac {d}{dx}}(3x-1)^{5}+(3x-1)^{5}{\frac {d}{dx}}(2x+1)^{3}\\[12pt]={}&(2x+1)^{3}\cdot 5(3x-1)^{4}\cdot 3+(3x-1)^{3}\cdot 3(2x+1)^{2}\cdot 2\\[12pt]={}&15(2x+1)^{3}(3x-1)^{4}+6(3x-1)^{3}(2x+1)^{2}\\[12pt]={}&(3x-1)^{3}(2x+1)^{2}{\Big (}15(2x+2)(3x-1)+6{\Big )}\\[12pt]={}&6(3x-1)^{3}(2x+1)^{2}{\Big (}15x^{2}+10x-4{\Big )}\\[12pt]={}&6(3x-1)^{3}(2x+1)^{2}\cdot 15\left(x-{\tfrac {5+{\sqrt {85}}}{15}}\right)\left(x-{\tfrac {5-{\sqrt {85}}}{15}}\right)\\[12pt]={}&90\cdot 3^{3}\cdot 2^{2}\cdot \left(x-{\frac {1}{3}}\right)^{3}\left(x+{\frac {1}{2}}\right)^{2}\left(x-{\tfrac {5+{\sqrt {85}}}{15}}\right)\left(x-{\tfrac {5-{\sqrt {85}}}{15}}\right)\end{aligned}}}
d
y
d
x
=
d
d
x
tan
x
1
+
cos
x
=
(
1
+
cos
x
)
d
d
x
tan
x
−
(
tan
x
)
d
d
x
(
1
+
cos
x
)
(
1
+
cos
x
)
2
=
(
1
+
cos
x
)
sec
2
x
+
(
tan
x
)
sin
x
(
1
+
cos
x
)
2
{\displaystyle {\begin{aligned}{}\\[8pt]{\frac {dy}{dx}}={}&{\frac {d}{dx}}\,{\frac {\tan x}{1+\cos x}}={\frac {(1+\cos x){\frac {d}{dx}}\tan x-(\tan x){\frac {d}{dx}}(1+\cos x)}{(1+\cos x)^{2}}}\\[12pt]={}&{\frac {(1+\cos x)\sec ^{2}x+(\tan x)\sin x}{(1+\cos x)^{2}}}\\[12pt]{}\end{aligned}}}
Problem #3 starts here:
x
e
y
=
y
sin
x
x
d
d
x
e
y
+
e
y
d
d
x
x
=
y
(
d
d
x
sin
x
)
+
(
sin
x
)
d
y
d
x
x
e
y
d
y
d
x
+
e
y
=
y
cos
x
+
(
sin
x
)
d
y
d
x
x
e
y
d
y
d
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−
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sin
x
)
d
y
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=
y
cos
x
−
e
y
(
x
e
y
−
sin
x
)
d
y
d
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=
y
cos
x
−
e
y
d
y
d
x
=
y
cos
x
−
e
y
x
e
y
−
sin
x
{\displaystyle {\begin{aligned}&xe^{y}=y\sin x\\[12pt]&x{\frac {d}{dx}}e^{y}+e^{y}{\frac {d}{dx}}x=y\left({\frac {d}{dx}}\sin x\right)+(\sin x){\frac {dy}{dx}}\\[12pt]&xe^{y}{\frac {dy}{dx}}+e^{y}=y\cos x+(\sin x){\frac {dy}{dx}}\\[12pt]&xe^{y}{\frac {dy}{dx}}-(\sin x){\frac {dy}{dx}}=y\cos x-e^{y}\\[12pt]&(xe^{y}-\sin x){\frac {dy}{dx}}=y\cos x-e^{y}\\[12pt]&{\frac {dy}{dx}}={\frac {y\cos x-e^{y}}{xe^{y}-\sin x}}\\[12pt]{}\end{aligned}}}
Problem #4 starts here:
y
=
(
arcsin
(
2
x
)
)
2
+
cot
(
csc
x
)
d
y
d
x
=
2
(
arcsin
(
2
x
)
)
⋅
d
d
x
arcsin
(
2
x
)
−
csc
2
(
csc
x
)
⋅
(
−
csc
x
cot
x
)
=
2
(
arcsin
(
2
x
)
)
⋅
1
1
−
(
2
x
)
2
⋅
d
d
x
(
2
x
)
+
csc
2
(
csc
x
)
(
csc
x
cot
x
)
=
2
(
arcsin
(
2
x
)
)
⋅
1
1
−
(
2
x
)
2
⋅
2
+
csc
2
(
csc
x
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csc
x
cot
x
)
=
2
(
arcsin
(
2
x
)
)
⋅
1
1
−
(
2
x
)
2
⋅
2
+
(
csc
(
csc
x
)
)
2
(
csc
x
cot
x
)
=
4
arcsin
(
2
x
)
1
−
(
2
x
)
2
+
(
csc
(
csc
x
)
)
2
(
csc
x
cot
x
)
{\displaystyle {\begin{aligned}y={}&(\arcsin(2x))^{2}+\cot(\csc x)\\[12pt]{\frac {dy}{dx}}={}&2(\arcsin(2x))\cdot {\frac {d}{dx}}\arcsin(2x)-\csc ^{2}(\csc x)\cdot (-\csc x\cot x)\\[12pt]={}&2(\arcsin(2x))\cdot {\frac {1}{\sqrt {1-(2x)^{2}}}}\cdot {\frac {d}{dx}}(2x)+\csc ^{2}(\csc x)(\csc x\cot x)\\[12pt]={}&2(\arcsin(2x))\cdot {\frac {1}{\sqrt {1-(2x)^{2}}}}\cdot 2+\csc ^{2}(\csc x)(\csc x\cot x)\\[12pt]={}&2(\arcsin(2x))\cdot {\frac {1}{\sqrt {1-(2x)^{2}}}}\cdot 2+{\big (}\csc(\csc x){\big )}^{2}(\csc x\cot x)\\[12pt]={}&{\frac {4\arcsin(2x)}{\sqrt {1-(2x)^{2}}}}+{\big (}\csc(\csc x){\big )}^{2}(\csc x\cot x)\\[12pt]{}\end{aligned}}}
Problem #5 starts here:
y
=
tan
(
10
x
)
+
10
tan
(
π
x
)
Recall that
d
d
x
10
x
=
10
x
⋅
log
e
10
d
y
d
x
=
sec
2
(
10
x
)
⋅
d
d
x
10
x
+
10
tan
(
π
x
)
⋅
d
d
x
tan
(
π
x
)
=
sec
2
(
10
x
)
⋅
10
x
⋅
ln
10
+
10
tan
(
π
x
)
⋅
sec
2
(
π
x
)
⋅
d
d
x
(
π
x
)
=
sec
2
(
10
x
)
⋅
10
x
⋅
ln
10
+
10
tan
(
π
x
)
⋅
sec
2
(
π
x
)
⋅
π
{\displaystyle {\begin{aligned}y={}&\tan(10^{x})+10^{\tan(\pi x)}\\[12pt]{\text{Recall that }}&{\frac {d}{dx}}10^{x}=10^{x}\cdot \log _{e}10\\[12pt]{\frac {dy}{dx}}={}&\sec ^{2}(10^{x})\cdot {\frac {d}{dx}}10^{x}+10^{\tan(\pi x)}\cdot {\frac {d}{dx}}\tan(\pi x)\\[12pt]={}&\sec ^{2}(10^{x})\cdot 10^{x}\cdot \ln 10+10^{\tan(\pi x)}\cdot \sec ^{2}(\pi x)\cdot {\frac {d}{dx}}(\pi x)\\[12pt]={}&\sec ^{2}(10^{x})\cdot 10^{x}\cdot \ln 10+10^{\tan(\pi x)}\cdot \sec ^{2}(\pi x)\cdot \pi \end{aligned}}}