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How do you implicitly differentiate #15=x^2+y^2#?

Answer 1

Ezra Adams

#(dy)/dx=-x/y#

#d/dx(15) =d/dx(x^2+y^2)#

#0=2x+2y(dy)/dx#

#2y(dy)/dx=-2x# ..... See foot note!

#(dy)/dx=-x/y#

#color(blue)(~~~~~~~~~~~"Foot Note" ~~~~~~~~~~~~~~~)#

#color(brown)("Let "u=y^2)#

#color(brown)((du)/dy =2y)#

#color(brown)("So "(du)/dx = (du)/dy *(dy)/dx)#

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Answer 2

Luke Alvarez

To implicitly differentiate (15 = x^2 + y^2), differentiate both sides of the equation with respect to (x).

(d/dx (15) = d/dx (x^2 + y^2))

(0 = 2x + 2y \frac{dy}{dx})

To solve for (\frac{dy}{dx}), rearrange the equation:

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

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Answer 3

Ezra Adams

To implicitly differentiate the equation (15 = x^2 + y^2), you differentiate both sides of the equation with respect to (x).

[ \frac{d}{dx}(15) = \frac{d}{dx}(x^2 + y^2) ]

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

[ \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.