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Differential Calculus

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Question : 66 of 69

Marks: +1, -0

if

e^{x y+x y}=e

\; \frac{d y}{d x} =

$-\;\frac{y}{x}$
$-\;\frac{1}{x y}$
$x y$
$\frac{x}{y}$

Solution:

Concept:

Simplify the exponential equation, then differentiate implicitly.

Explanation:

Given

e^{xy+xy} = e

.

Simplify exponent:

xy+xy = 2xy

. So

e^{2xy} = e

.

Thus

2xy = 1 \implies xy = \frac{1}{2}

.

Differentiate both sides with respect to

x

:

\frac{d}{dx}(xy) = \frac{d}{dx}\left(\frac{1}{2}\right)

.

Using product rule:

x \frac{dy}{dx} + y = 0

.

Solve for

\frac{dy}{dx}

:

x \frac{dy}{dx} = -y

⇒

\frac{dy}{dx} = -\frac{y}{x}

.

Answer:

-\frac{y}{x}

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