+0

0
191
4

Guest Jan 19, 2015

#1
+91491
+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
+91491
+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
+81077
+5

CPhill  Jan 19, 2015
#3
+91491
+5

Thanks Chris.

Melody  Jan 19, 2015
#4
+5

Ahh, thanks guys!

Guest Jan 21, 2015

### 5 Online Users

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