+0  
 
0
209
6
avatar+62 

$$f(x)=?$$

YehChi  Jan 13, 2015

Best Answer 

 #4
avatar+91432 
+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
Sort: 

6+0 Answers

 #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+18827 
+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+91432 
+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+91432 
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

8 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details