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How do you use implicit differentiation to find #(dy)/(dx)# given #sqrtx+sqrty=1#?

Answer 1

Emily Ammerman

# y' =- sqrty/(sqrtx) = 1 - 1/(sqrtx)#

#d/dx( sqrtx+sqrty=1)#

#implies 1/(2sqrtx)+1/(2 sqrty) y' =0#

# y' =- sqrty/(sqrtx)#

# y' =- (1 - sqrt x)/(sqrtx)#

# y' = 1 - 1/(sqrtx)#

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Answer 2

Charles Anctil

To find (\frac{dy}{dx}) using implicit differentiation for the equation (\sqrt{x}+\sqrt{y}=1), differentiate both sides of the equation with respect to (x), treating (y) as a function of (x). Then, solve for (\frac{dy}{dx}). Applying implicit differentiation, the derivative of (\sqrt{x}) is (\frac{1}{2\sqrt{x}}) and the derivative of (\sqrt{y}) is (\frac{1}{2\sqrt{y}}). Thus, (\frac{d}{dx}(\sqrt{x}+\sqrt{y})=\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}). Therefore, (\frac{dy}{dx}=-\frac{\sqrt{x}}{\sqrt{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.