If x^y.y^x=16, then the value of dy/dx at (2, 2) is

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MARKS Content Team · Accepted Answer

Correct answer: (A) -1. Since, x^y.y^x=16 ∴log_e⁡x^y+log_e⁡y^x=log_e⁡16 ⇒ylog_e⁡x+xlog_e⁡y=4log_e⁡2 Now, on differentiating both sides w.r.t. x, we get y/x+log_e⁡xdy/dx+x/ydy/dx+log_e⁡y.1=0 ⇒dy/dx=-(log_e⁡y+y)/xlog_e⁡x+x/y ∴dy_(dx(2,2))=-((log_e⁡2+1))/((log_e⁡2+1))=-1

If $x^{y} . y^{x} = 16$ , then the value of $\frac{d y}{d x}$ at $(2, 2)$ is

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More Differentiation Questions

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

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More Differentiation Questions

If $x^{y} . y^{x} = 16$ , then the value of $\frac{d y}{d x}$ at $(2, 2)$ is