If xy ⋅ yx=16 then ((d y/d x))(2,2) is equal to

Question

If xy ⋅ yx=16 then ((d y/d x))(2,2) is equal to (A) 1 (B) -1 (C) 0 (D)  ln 2+1

Tardigrade Admin · Accepted Answer

Taking log on both sides, y ln x + x ln y = ln 16 differentiating w.r.t. x, we get (y/x)+ ln x ⋅ y prime+(x/y) ⋅ y prime+ ln y=0 text putting x=y=2 1+ ln 2 ⋅ y prime+y prime+ ln 2=0 y prime=-1

Q. If $x^{y} \cdot y^{x} = 16$ then $(\frac{d y}{d x})_{(2, 2)}$ is equal to

1

-1

0

$ln 2 + 1$

Taking log on both sides,
$y ln x + x ln y = ln 16$
differentiating w.r.t. $x$ , we get
$\frac{y}{x} + ln x \cdot y^{'} + \frac{x}{y} \cdot y^{'} + ln y = 0$
$putting x = y = 2$
$1 + ln 2 \cdot y^{'} + y^{'} + ln 2 = 0$
$y^{'} = - 1$

Q. If xy⋅yx=16 then (dxdy​)(2,2)​ is equal to

1

-1

0

ln2+1

Taking log on both sides, ylnx+xlny=ln16 differentiating w.r.t. x, we get xy​+lnx⋅y′+yx​⋅y′+lny=0 putting x=y=2 1+ln2⋅y′+y′+ln2=0 y′=−1

Q. If $x^{y} \cdot y^{x} = 16$ then $(\frac{d y}{d x})_{(2, 2)}$ is equal to

$ln 2 + 1$

Taking log on both sides,
$y ln x + x ln y = ln 16$
differentiating w.r.t. $x$ , we get
$\frac{y}{x} + ln x \cdot y^{'} + \frac{x}{y} \cdot y^{'} + ln y = 0$
$putting x = y = 2$
$1 + ln 2 \cdot y^{'} + y^{'} + ln 2 = 0$
$y^{'} = - 1$