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Please help me differentiate in terms of x and y for sin(x+y)=cosy

Guest Jan 19, 2015

Best Answer 

 #1
avatar+93289 
+10

PLEASE NOTE :  I have no idea if this is correct - someone please check   

sin(x+y)=cosy

differentiate y in terms of x

 

$$\begin{array}{rll}
sin(x+y)&=&cosy\\\\
cos(x+y)[1+\frac{dy}{dx}]&=&-(siny)\frac{dy}{dx}\\\\
cos(x+y)+cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&0\\\\
cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&-cos(x+y)\\\\
\frac{dy}{dx}[cos(x+y)+(siny)]&=&-cos(x+y)\\\\
\frac{dy}{dx}&=&\frac{-cos(x+y)}{cos(x+y)+(siny)}\\\\
\end{array}$$

 

If that is correct then i guess

 

$$\frac{dx}{dy}&=&\frac{cos(x+y)+(siny)}{-cos(x+y)}\\\\$$

Melody  Jan 19, 2015
 #1
avatar+93289 
+10
Best Answer

PLEASE NOTE :  I have no idea if this is correct - someone please check   

sin(x+y)=cosy

differentiate y in terms of x

 

$$\begin{array}{rll}
sin(x+y)&=&cosy\\\\
cos(x+y)[1+\frac{dy}{dx}]&=&-(siny)\frac{dy}{dx}\\\\
cos(x+y)+cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&0\\\\
cos(x+y)\frac{dy}{dx}+(siny)\frac{dy}{dx}&=&-cos(x+y)\\\\
\frac{dy}{dx}[cos(x+y)+(siny)]&=&-cos(x+y)\\\\
\frac{dy}{dx}&=&\frac{-cos(x+y)}{cos(x+y)+(siny)}\\\\
\end{array}$$

 

If that is correct then i guess

 

$$\frac{dx}{dy}&=&\frac{cos(x+y)+(siny)}{-cos(x+y)}\\\\$$

Melody  Jan 19, 2015
 #2
avatar+88775 
+5

Both answers are correct, Melody....

 

CPhill  Jan 19, 2015
 #3
avatar+93289 
+5

Thanks Chris.   

Melody  Jan 19, 2015
 #4
avatar
+5

Ahh, thanks guys!

Guest Jan 21, 2015

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