Question

If exy = 4, then what is dy over dx at the point (1, ln4)?

Answer 1

the value of dy/dx at (1, ln4) = ln(1/4)

given, equation of curve, e^xy = 4

and we have to find dy/dx at the point (1, ln4)

first differentiate curve with respect to x.

d(e^xy)/dx = d(4)/dx = 0

⇒e^(xy) [x dy/dx + y dx/dx ] = 0

⇒e^(xy) [x dy/dx + y ] = 0

⇒dy/dx = -y/x

now at (1, ln4) , dy/dx = -ln4/1

⇒dy/dx = -ln4 = ln(4)^-1 = ln(1/4)

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the formula on derivatives that :

If u and v are two differentiable function then

$\displaystyle \: \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

GIVEN

${e}^{xy} = 4$

TO DETERMINE

$\displaystyle \: \sf{The \: value \: of \: \frac{dy}{dx} \: at \: (1 \: , \: ln4) \: \: \: i.e. }$

$\displaystyle \: \bigg[ \: \frac{dy}{dx} \bigg]_{ (1, \: ln4)}$

EVALUATION

Here

${e}^{xy} = 4$

Taking logarithm in both sides we get

$ln( {e}^{xy} )=ln( 4)$

$\implies \: xy \: ln(e) = ln \: 4$

$\implies \: xy \: = ln \: 4$

Differentiating both sides with respect to x we get

$\displaystyle \: y + x \frac{dy}{dx} = 0$

$\sf{Putting \: x = 1 \: , \: y = ln 4 \: we \: get }$

$ln \: 4 + 1 \times \displaystyle \: \bigg[ \: \frac{dy}{dx} \bigg]_{ (1, \: ln4)} = 0$

$\implies \: \displaystyle \: \bigg[ \: \frac{dy}{dx} \bigg]_{ (1, \: ln4)} = - ln \: 4$

RESULT

The required answer is

$\boxed{ \: \displaystyle \: \bigg[ \: \frac{dy}{dx} \bigg]_{ (1, \: ln4)} = - ln \: 4\: \: }$