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How do you find #dy/dx# by implicit differentiation of #x^2+y^2=36#?

Answer 1

Elijah Abram

#dy/dx = -y/x#

#d/dx(x^2+y^2) = d/dx36#

#=> d/dxx^2 + d/dxy^2 = 0#

#=> 2x + 2y(d/dxy) = 0# (Applying the chain rule with #y# treated as a function of #x#)

#=> 2x+2ydy/dx = 0#

#=> 2ydy/dx = -2x#

#:. dy/dx = -x/y#

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Answer 2

Isaiah Allen

To find ( \frac{dy}{dx} ) by implicit differentiation of ( x^2 + y^2 = 36 ), follow these steps:

Differentiate both sides of the equation with respect to ( x ).
Use the chain rule for differentiating terms involving ( y ).
Solve for ( \frac{dy}{dx} ) in terms of ( x ) and ( y ).

Starting with the given equation:

[ x^2 + y^2 = 36 ]

Differentiate both sides with respect to ( x ):

[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(36) ]

[ 2x + 2y \frac{dy}{dx} = 0 ]

Isolate ( \frac{dy}{dx} ) to solve for it:

[ 2y \frac{dy}{dx} = -2x ]

[ \frac{dy}{dx} = \frac{-2x}{2y} ]

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

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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.