If xy - yx - ab then show that dydx - - - yxy - 1 - yxlog yxylog x - xyx - 1-

x^y+y^x=a^b e^y ln x+e^x ln y=a^b Differentiating w.r.t x x^y[dydxln x+yx]+y^x[ln y+xydydx]=0 dydx[x^yln x+y^x 1x]= [y^xln y+yx^y 1] ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

If xy + yx ...

Question

If $x^{y} + y^{x} = a^{b}$ then show that $\frac{d y}{d x} = - [\frac{y x^{y - 1} + y^{x} log y}{x^{y} log x + x y^{x - 1}}]$

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Solution

$x^{y} + y^{x} = a^{b}$

$e^{y l n x} + e^{x l n y} = a^{b}$

Differentiating w.r.t $x$

$x^{y} [\frac{d y}{d x} l n x + \frac{y}{x}] + y^{x} [l n y + \frac{x}{y} \frac{d y}{d x}] = 0$

$\frac{d y}{d x} [x^{y} l n x + y^{x - 1} x] = - [y^{x} l n y + y x^{y - 1}]$

$\frac{d y}{d x} = - \frac{- y^{x} l n y + y x^{y - 1}}{x^{y} l n x + x y^{x - 1}}$ .

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If xy+yx=ab then show that dydx=−[yxy−1+yxlogyxylogx+xyx−1]

xy+yx=abeylnx+exlny=abDifferentiating w.r.t xxy[dydxlnx+yx]+yx[lny+xydydx]=0dydx[xylnx+yx−1x]=−[yxlny+yxy−1]dydx=−−yxlny+yxy−1xylnx+xyx−1.

If $x^{y} + y^{x} = a^{b}$ then show that $\frac{d y}{d x} = - [\frac{y x^{y - 1} + y^{x} log y}{x^{y} log x + x y^{x - 1}}]$