Question

find dy/dx when:xy log(x+y)=1

Answer 1

the answer is above....

Answer 2

Answer:

Given:

xy log(x+y)=1

To find: $\frac{\mathrm{d} y}{\mathrm{d} x}$

We use Product rule,

(x.y.z)' = x.y.z' + x.z.y' + y.z.x'

Consider,

xy log(x+y)=1

Derivate both sides,

$\frac{\mathrm{d}(x.y.log(x+y))}{\mathrm{d}x}=\frac{\mathrm{d}(1)}{\mathrm{d} x}$

$x.y.\frac{\mathrm{d}(log(x+y))}{\mathrm{d}x}+x.log(x+y).\frac{\mathrm{d}(y)}{\mathrm{d}x}+y.log(x+y).\frac{\mathrm{d}(x)}{\mathrm{d}x}=0$

$x.y.\frac{1}{x+y}\frac{\mathrm{d}(x+y)}{\mathrm{d}x}+x.log(x+y).\frac{\mathrm{d}y}{\mathrm{d}x}+y.log(x+y)=0$

$x.y.\frac{1}{x+y}(\frac{\mathrm{d}(x))}{\mathrm{d}x}+\frac{\mathrm{d} (y)}{\mathrm{d} x})+x.log(x+y).\frac{\mathrm{d}y}{\mathrm{d}x}=-y.log(x+y)$

$\frac{xy}{x+y}+\frac{xy}{x+y}.\frac{\mathrm{d} (y)}{\mathrm{d} x}+x.log(x+y).\frac{\mathrm{d}y}{\mathrm{d}x}=-y.log(x+y)$

$(\frac{xy}{x+y}+x.log(x+y))\frac{\mathrm{d}y}{\mathrm{d}x}=-y.log(x+y)-\frac{xy}{x+y}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-y.log(x+y)-\frac{xy}{x+y}}{\frac{xy}{x+y}+x.log(x+y)}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-y(x+y).log(x+y)-xy}{xy+x(x+y)log(x+y)}$