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Math Help - find dy/dx by implicit differentiation

  1. #1
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    find dy/dx by implicit differentiation

    X^(1/2)+y^(1/2)=16 can some one show me step by step because the teacher wants it written out so im trying to get it on the homework.
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  2. #2
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    The equation \sqrt{x}+\sqrt{y}=16 states that the functions f(x)=\sqrt{x}+\sqrt{y} and g(x)=16 are equal everywhere. Because they are equal, they have equal derivatives:

    \frac{d}{dx}(\sqrt{x})+\frac{d}{dx}(\sqrt{y})=\fra  c{d}{dx}(16).

    With the Chain Rule, we can now find y' in terms of x and y.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Nightasylum View Post
    X^(1/2)+y^(1/2)=16 can some one show me step by step because the teacher wants it written out so im trying to get it on the homework.
    \sqrt x  + \sqrt y  = 16 \Leftrightarrow \sqrt y  = 16 - \sqrt x  \Rightarrow

    \Rightarrow \frac{d}<br />
{{dx}}\sqrt y  = \frac{d}<br />
{{dx}}\left( {16 - \sqrt x } \right) \Leftrightarrow \frac{{dy}}<br />
{{dx}} \cdot \frac{1}<br />
{{2\sqrt y }} =  - \frac{1}<br />
{{2\sqrt x }} \Leftrightarrow

    \Leftrightarrow \frac{{dy}}<br />
{{dx}} =  - \frac{{\sqrt y }}<br />
{{\sqrt x }} =  - \frac{{16 - \sqrt x }}<br />
{{\sqrt x }} = 1 - \frac{{16}}<br />
{{\sqrt x }}.
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