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How do you find dy/dx by implicit differentiation for #sinx + 2cos2y = 1#?

Answer 1

William Abdalla

#dy/dx=cos(x)/(4sin(2y))#

Differentiate every term with respect to #x#

#cos(x)-4sin(2y)dy/dx=0#

isolate the term containing #dy/dx# by adding

#4sin(2y)dy/dx# to both sides

#4sin(2y)dy/dx=cos(x)#

Divide both sides by #4sin(2y)#

#dy/dx=cos(x)/(4sin(2y))#

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Answer 2

Evelyn Agard

To find ( \frac{dy}{dx} ) by implicit differentiation for the equation ( \sin(x) + 2\cos(2y) = 1 ), follow these steps:

Differentiate both sides of the equation with respect to ( x ).
Use the chain rule and product rule where necessary.
Solve the resulting equation for ( \frac{dy}{dx} ).

Differentiating both sides of the equation:

[ \frac{d}{dx}(\sin(x)) + \frac{d}{dx}(2\cos(2y)) = \frac{d}{dx}(1) ]

[ \cos(x) - 4\sin(2y) \frac{dy}{dx} = 0 ]

Solve for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = \frac{\cos(x)}{4\sin(2y)} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.