+0

0
188
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Guest Jan 19, 2015

#1
+91451
+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
Sort:

#1
+91451
+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
#2
+81004
+5

CPhill  Jan 19, 2015
#3
+91451
+5

Thanks Chris.

Melody  Jan 19, 2015
#4
+5

Ahh, thanks guys!

Guest Jan 21, 2015

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