If xy.yx=16,then dydx at (2,2) is

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Question

If $x^{y} . y^{x} = 16, t h e n \frac{d y}{d x} a t (2, 2)$ is

Options:

A . - 1

B . 0

C . 1

D . None of these

Answer: Option A
:
A

x^{y} . y^{x} = 16

∴ l o g_{e} x^{y} + l o g_{e} y^{x} = l o g_{e} 16

\Rightarrow y l o g_{e} x + x l o g_{e} y = 4 l o g_{e} 2

Now, differentiating both sides w.r.t.x

\frac{y}{x} + l o g_{e} x \frac{d y}{d x} + \frac{x}{y} \frac{d y}{d x} + l o g_{e} y .1 = 0

∴ \frac{d y}{d x} = - \frac{(l o g_{e} y + \frac{y}{x})}{(l o g_{e} x + \frac{x}{y})}

∴ \frac{d y}{d x} |_{(2, 2)} = - \frac{(l o g_{e} 2 + 1)}{(l o g_{e} 2 + 1)} = - 1

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