Question

Differentiate this xylog (x+y)=1

Answer 1

I send you solution look

Answer 2

$\frac{dy}{dx} = - \dfrac{[y \log (x + y) + \frac{xy}{x + y} ]}{[x\log (x + y) + \frac{xy}{x + y} ]}$

Step-by-step explanation:

We have to differentiate the equation xy log(x + y) = 1  ........... (1) and hence find the value of $\frac{dy}{dx}$.

Now, differentiating both sides of equation (1) with respect to x we get

$\frac{d}{dx}[xy \log (x + y)] = 0$

⇒ $\log (x+ y) \frac{d}{dx} (xy) + xy \frac{d}{dx} [\log (x + y)] = 0$

⇒ $\log (x + y)[y + x\frac{dy}{dx}] + \frac{xy}{x + y}[1 + \frac{dy}{dx} ] = 0$ {Since, differentiation of log x is 1/x}

⇒ $[x\log (x + y) + \frac{xy}{x + y}]\frac{dy}{dx} = - [y \log (x + y) + \frac{xy}{x + y} ]$

⇒ $\frac{dy}{dx} = - \dfrac{[y \log (x + y) + \frac{xy}{x + y} ]}{[x\log (x + y) + \frac{xy}{x + y} ]}$  (Answer)