Question

Find dy/dx when ysinx=xcosy
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Answer 1

Your answer is in the attachment

Answer 2

$\bf \: \frac{dy}{dx} = \frac{cos \: y -y \: cos \: x}{sin \: x + x \: sin \: y}$

Given:

• $y \: sin \: x= x \: cos \: y$

To find:

• Find $\frac{dy}{dx}$

Solution:

Concept to be used:

• Apply chain rule.

Step 1:

Do differentiation with respect to x .

Remember that it is implicit differentiation.

$y\frac{d}{dx}(sin \: x) + sin \: x \: \frac{dy}{dx} =x\frac{d}{dx}(cos\: y) + cos \: y\: \frac{dx}{dx}$

$y \: cos \: x + sin \: x \: \frac{dy}{dx} = - x \: sin \: y \: \frac{dy}{dx} + cos \: y$

Step 2:

Simply the equation.

Take the terms containing dy/dx to LHS.

$sin \: x \: \frac{dy}{dx} + x \: sin \: y \: \frac{dy}{dx} = cos \: y - y \: cos \: x$

$(sin \: x + x \: sin \: y) \: \frac{dy}{dx} = cos \: y - y \: cos \: x$

$\frac{dy}{dx} = \frac{cos \: y - y \: cos \: x}{sin \: x + x \: sin \: y}$

Thus,

$\bf \: \frac{dy}{dx} = \frac{cos \: y - y \: cos \: x}{sin \: x + x \: sin \: y}$

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