If xy- yx-16- then d y-d x at 2-2 isA- -1B- 0C- 1D- None of these

The correct option is A 1x^y.y^x = 16 ∴ log_ex^y + log_e y^x = log_e 16 ⇒ y log_e x +x log_e y = 4 log_e 2 Now, differentiating both sides w.r.t.x y/x + log ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

If xy· yx=16,...

Question

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

- 1

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None of these

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Solution

The correct option is A - 1
$x^{y} . y^{x} = 16$
$∴ l o g_{e} x^{y} + l o g_{e} y^{x} = l o g_{e} 16$
$\Rightarrow y l o g_{e} x + x l o g_{e} y = 4 l o g_{e} 2$
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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If xy.yx=16,then dydx at (2,2) is

The correct option is A - 1xy.yx=16 ∴ logexy+loge yx=loge 16 ⇒y loge x+x loge y=4 loge 2 Now, differentiating both sides w.r.t.x yx+loge x dydx+xydydx+loge y.1=0 ∴dydx=−(loge y+yx)(loge x+xy) ∴ dydx|(2,2)=−(loge 2+1)(loge 2+1)=−1

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