Question

If xy + yx = ab, find dy/dx.

Answer 1

Step-by-step explanation:

The given function is xy - yx = ab

Let xy = u and yx = v

Then , the function becomes u - v = ab

dudx-dvdx=0 .....(1)

u = xy

⇒ log u = log(xy)

⇒ log u = y log x

Differentiating both sides with respect to x, we obtain

1ududx=logxdydx+y.ddx(logx)

⇒

dudx=[logxdydx+y.1x]

⇒

dudx=xy(logx dydx+yx) ...(2)

Answer 2

Answer :-

$› \: \frac{dy}{dx} = \frac{ - ( {x}^{y}. \frac{y}{x} + \: {y}^{x}. \: log \: y) }{( {x}^{y} . \: log \: x \: + {y}^{x}. \frac{x}{y}) }$

Step-by-step explanation :-

$› \: {x}^{y} (y \: log \: x \: + \frac{y}{x} ) \: + \: {y}^{x} (log \: y \: + \frac{x}{y} \: .y \: ) \: = 0$

$› \: y \: ( {x}^{y} \: log \: x \: + \: {y}^{x} \: . \frac{x}{y})= - {x}^{y} ( \frac{y}{x} - {y}^{x} \: log \: x \: )$

$› \: \frac{dy}{dx} = \frac{- ( {x}^{y}. \frac{y}{x} + {y}^{x}. log \: y)}{( {x}^{y}.log \: x \: + {y}^{x}. \frac{x}{y})}$

Hope it helps!