# `x sin(y) + y sin(x) = 1` Find `(dy/dx)` by implicit differentiation.

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### 2 Answers

Differentiate both sides of the equation with respect to x.

`d/dx(xsin(y)) + d/dx(ysin(x)) = d/dx 1`

Each of these terms needs to be differentiated with the product rule.

`x d/dx(siny) + sin(y)d/dx (x) + y d/dx (sinx) + sinx d/dx (y) = 0`

`x(cosy y') + sin(y)1 + y(cos(x)) + sinx y' = 0`

`xy' cosy + siny + ycosx + y' sinx = 0`

Now solve for y'

`(xcosy + sinx)y' = -siny - ycosx`

`y' = (-siny - ycosx)/(xcosy + sinx)`

*Note:- 1) If y = cosx ; then dy/dx = -sinx *

*2) If y = k ; where k = constant ; then dy/dx = 0*

*3) If y = u*v ; where both u & v are functions of 'x' , then*

*dy/dx = u*(dv/dx) + v*(du/dx)*

*4) If y = sinx ; then dy/dx = cosx*

Now, the given function is :-

(xy)*sin(x)*sin(y) = 1

Differentiating both sides w.r.t 'x' we get

y*sin(x)*sin(y) + [x*sin(x)*sin(y)]*(dy/dx) + (xy)*cos(x)*sin(y) + [(xy)*sin(x)*cos(y)]*(dy/dx) = 0

dy/dx = -[{y*sin(x)*sin(y)} + {(xy)*cos(x)*sin(y)}]/[{(xy)*sin(x)*cos(y)} + {x*sin(x)*sin(y)}]