+0  
 
0
425
6
avatar+62 

$$f(x)=?$$

YehChi  Jan 13, 2015

Best Answer 

 #4
avatar+94105 
+10

I am going to use quotient rule.

$$\\u=(x+1)^{0.5}\\
u'=0.5*(x+1)^{-0.5}\\\\\\
v=(x+2)[sin(3x+2)]^2\\\\
v'=1*[sin(3x+2)]^2\;\;+\;\;2(sin(3x+2))*cos(3x+2)*3\\\\
v'=sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)\\\\\\
\frac{dy}{dx}=(e+1)^3\frac{vu'+uv'}{v^2}\\\\
vu'=(x+2)sin^2(3x+2)*0.5*(x+1)^{-0.5}\\\\
vu'=\frac{0.5(x+2)sin^2(3x+2)}{(x+1)^{0.5}}\\\\\\
uv'=(x+1)^{0.5}*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)\\\\
uv'=\frac{(x+1)*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}$$

 

$$\\vu'=\frac{0.5(x+2)sin^2(3x+2)}{(x+1)^{0.5}}\\\\\\
uv'=\frac{(x+1)*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}\\\\
vu'+uv'=\frac{0.5(x+2)sin^2(3x+2)+(x+1)sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}$$

 

$$\\\frac{vu'+uv'}{v^2}=\frac{[0.5(x+2)sin^2(3x+2)]+[(x+1)sin^2(3x+2)]\;\;+\;\;[6sin(3x+2)cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^4(3x+2)(x+1)^{0.5}}
\\\\\frac{vu'+uv'}{v^2}=\frac{[0.5(x+2)sin(3x+2)]+[(x+1)sin(3x+2)]\;\;+\;\;[6cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^3(3x+2)(x+1)^{0.5}}$$

 

$$\\\frac{dy}{dx}=(e+1)^3\times \frac{[0.5(x+2)sin(3x+2)]+[(x+1)sin(3x+2)]\;\;+\;\;[6cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^3(3x+2)(x+1)^{0.5}}\\\\$$

 

 

There is probably only about 1000 mistakes in there.

Let's see what Heureka found   ( I saw Heureka's pop-up)

Melody  Jan 13, 2015
 #1
avatar
+5

Is (3x+2)^2     really in degrees?

radians are used for calculus.

Guest Jan 13, 2015
 #2
avatar+62 
0

I replaced with pictures.

YehChi  Jan 13, 2015
 #3
avatar+20679 
+10

$$\small{\text{
$
f(x)=\dfrac{ (e+1)^3\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \\
\\
$
}}
\samll{\text{
$ \qquad \textcolor[rgb]{1,0,0}{f'(x) = ?} $
}}$\\\\$
\small{\text{
$
f(x)=\dfrac{ (e+1)^3\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } = (e+1)^3 \left(
\sqrt{x+1} *\frac{1}{x+2} * \frac{1}{\sin{(3x+2)} } * \frac{1}{\sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=(e+1)^3\dfrac{\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { ( \sqrt{x+1} )' } { \sqrt{x+1} }
-\dfrac { ( x+2 )' } { x+2 }
-\dfrac { ( \sin{(3x+2)} )' } { \sin{(3x+2)} }
-\dfrac { ( \sin{(3x+2)} )' } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=(e+1)^3\dfrac{\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { 1 } { 2\sqrt{x+1}\sqrt{x+1} }
-\dfrac { 1 } { x+2 }
-\dfrac { 3\cos{(3x+2)} } { \sin{(3x+2)} }
-\dfrac { 3\cos{(3x+2)} } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=(e+1)^3\dfrac{\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { 1 } { 2\sqrt{x+1}\sqrt{x+1} }
-\dfrac { 1 } { x+2 }
-\dfrac { 2*3\cos{(3x+2)} } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=\dfrac{ (e+1)^3\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { 1 } { 2( x+1 ) }
-\dfrac { 1 } { x+2 }
-\dfrac { 6\cos{(3x+2)} } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
P.S.
$
(uv)' = uv\left( \frac{u'}{u} + \frac{v'}{v} \right)
$
and
$
(\frac{u}{v})' = \frac{u}{v}\left( \frac{u'}{u} - \frac{v'}{v} \right)
$
and
$
(\frac{u}{v*w})' = \frac{u}{v*w}\left( \frac{u'}{u} - \frac{v'}{v} - \frac{w'}{w} \right)
$
and
$
(\frac{u}{v*w*w})' = \frac{u}{v*w*w}\left( \frac{u'}{u} - \frac{v'}{v} - \frac{w'}{w} - \frac{w'}{w} \right)
$
}}$$

heureka  Jan 13, 2015
 #4
avatar+94105 
+10
Best Answer

I am going to use quotient rule.

$$\\u=(x+1)^{0.5}\\
u'=0.5*(x+1)^{-0.5}\\\\\\
v=(x+2)[sin(3x+2)]^2\\\\
v'=1*[sin(3x+2)]^2\;\;+\;\;2(sin(3x+2))*cos(3x+2)*3\\\\
v'=sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)\\\\\\
\frac{dy}{dx}=(e+1)^3\frac{vu'+uv'}{v^2}\\\\
vu'=(x+2)sin^2(3x+2)*0.5*(x+1)^{-0.5}\\\\
vu'=\frac{0.5(x+2)sin^2(3x+2)}{(x+1)^{0.5}}\\\\\\
uv'=(x+1)^{0.5}*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)\\\\
uv'=\frac{(x+1)*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}$$

 

$$\\vu'=\frac{0.5(x+2)sin^2(3x+2)}{(x+1)^{0.5}}\\\\\\
uv'=\frac{(x+1)*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}\\\\
vu'+uv'=\frac{0.5(x+2)sin^2(3x+2)+(x+1)sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}$$

 

$$\\\frac{vu'+uv'}{v^2}=\frac{[0.5(x+2)sin^2(3x+2)]+[(x+1)sin^2(3x+2)]\;\;+\;\;[6sin(3x+2)cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^4(3x+2)(x+1)^{0.5}}
\\\\\frac{vu'+uv'}{v^2}=\frac{[0.5(x+2)sin(3x+2)]+[(x+1)sin(3x+2)]\;\;+\;\;[6cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^3(3x+2)(x+1)^{0.5}}$$

 

$$\\\frac{dy}{dx}=(e+1)^3\times \frac{[0.5(x+2)sin(3x+2)]+[(x+1)sin(3x+2)]\;\;+\;\;[6cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^3(3x+2)(x+1)^{0.5}}\\\\$$

 

 

There is probably only about 1000 mistakes in there.

Let's see what Heureka found   ( I saw Heureka's pop-up)

Melody  Jan 13, 2015
 #5
avatar+62 
+5

Melody,You are too complicated...

$$lny=ln(e+1)^3+\(\frac{1}{2}(x+1)-ln(x+2)-2ln(sin(3x+2))

\(\frac{y'}{y}=0+\(\frac{1}{2(x+1)}-\(\frac{1}{x+2}-\(\frac{2cos(3x+2)3}{sin(3x+2)}

y'=y[\(\frac{1}{2(x+1)}-\(\frac{1}{x+2}-\(\frac{6cos(3x+2)}{sin(3x+2)}]

y'=\(\frac{(e+1)^3\sqrt{x+1}}{(x+2)sin^2(3x+2)}[\(\frac{1}{2(x+1)}-\(\frac{1}{x+2}-6cot(3x+2)]$$

YehChi  Jan 13, 2015
 #6
avatar+94105 
0

Thanks YehChi, I shall have to look at it when I am fresher.    

I just made it an entire fraction. 

Melody  Jan 13, 2015

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