If y^x + x^y = a^b , then dy dx = - Vidyadip

MCQ

If ${y^x} + {x^y} = {a^b}$, then ${{dy} \over {dx}} = $

✓
$ - {{y{x^{y - 1}} + {y^x}\log y} \over {x{y^{x - 1}} + {x^y}\log x}}$
B
${{y{x^{y - 1}} + {y^x}\log y} \over {x{y^{x - 1}} + {x^y}\log x}}$
C
$ - {{y{x^{y - 1}} + {y^x}} \over {x{y^{x - 1}} + {x^y}l}}$
D
${{y{x^{y - 1}} + {y^x}} \over {x{y^{x - 1}} + {x^y}}}$

✓

Answer

Correct option: A.

$ - {{y{x^{y - 1}} + {y^x}\log y} \over {x{y^{x - 1}} + {x^y}\log x}}$

a
(a) ${x^y} + {y^x} = {a^b}$;

Let ${x^y} = u$ and ${y^x} = v$

==> $u + v = {a^b}$

==>$\frac{{du}}{{dx}} + \frac{{dv}}{{dx}} = 0$

Now differentiating both by taking their $\log $ separately

$\frac{{du}}{{dx}} = {x^y}\left( {\frac{y}{x} + \frac{{dy}}{{dx}}\log x} \right)$ …..$(i)$

and $\frac{{dv}}{{dx}} = {y^x}\left( {\log y + \frac{x}{y}.\frac{{dy}}{{dx}}} \right)$ …..$(ii)$

Therefore, by $(i)$ and $(ii),$

$\frac{{dy}}{{dx}} = - \frac{{y{x^{y - 1}} + {y^x}\log y}}{{{x^y}\log x + x{y^{x - 1}}}}$.

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