Differential Help (2)

[nil101]

Is this answer correct please?

Differentiate the following wrt x:

\(\displaystyle \L y = \frac{{\ln (\sin ^2 3x)}}{{(1 - x)}}\)



\(\displaystyle \L u = \ln (\sin ^2 3x){\rm }u' = \left( {\frac{{3\sin 6x}}{{\sin ^2 3x}}} \right)\)

\(\displaystyle \L v = 1 - x{\rm }v' = - 1\)


\(\displaystyle \L \frac{{dy}}{{dx}} = \frac{{\left( {1 - x} \right)\left( {\frac{{3\sin 6x}}{{\sin ^2 3x}}} \right) + \ln (\sin ^2 3x)}}{{\left( {1 - x} \right)^2 }}\)

\(\displaystyle \L = \frac{{\left( {1 - x} \right)\left( {3\sin 6x} \right) + \ln (\sin ^2 3x)^2 }}{{\left( {1 - x} \right)^2 }}\)

Thanks for checking.

 

[soroban]

Hello, nil101!

You started fine, then made a couple of errors . . .

Differentiate: \(\displaystyle \L\,y \:= \:\frac{\ln (\sin ^2 3x)}{(1 - x)}\)

\(\displaystyle u\:=\:\ln(\sin^23x)\;\;\Rightarrow\;\;u'\:=\:\frac{1}{\sin^23x}\,\cdot\,2\,\cdot\,\sin3x\,\cdot\,\cos3x\,\cdot\,3 \:= \:\frac{6\cos3x}{\sin3x}\,=\,6\cdot\cot3x\)

\(\displaystyle v\:=\:1\,-\,x\;\;\Rightarrow\;\;v'\,=\,-1\)


\(\displaystyle \L\frac{dy}{dx}\;=\;\frac{(1\,-\,x)\cdot6\cdot\cot3x\,-\,(-1)\ln(\sin^23x)}{(1\,-\,x)^2}\)

\(\displaystyle \L\frac{dy}{dx}\;=\;\frac{6(1\,-\,x)\cot3x\,+\,\ln(\sin^23x)}{(1\,-\,x)^2}\)

 

[skeeter]

make life a little easier ... use the laws of logs before differentiating

\(\displaystyle y = \frac{ln[sin^2(3x)]}{1-x} = \frac{2ln[sin(3x)]}{1-x}\)

\(\displaystyle \frac{dy}{dx} = \frac{(1-x)6cot(3x) + 2ln[sin(3x)]}{(1-x)^2}\)

 

[nil101]

Thanks very much guys for your help!