# Thread: find dy/dx by implicit differentiation

1. ## 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.

2. The equation $\displaystyle \sqrt{x}+\sqrt{y}=16$ states that the functions $\displaystyle f(x)=\sqrt{x}+\sqrt{y}$ and $\displaystyle g(x)=16$ are equal everywhere. Because they are equal, they have equal derivatives:

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

With the Chain Rule, we can now find $\displaystyle y'$ in terms of $\displaystyle x$ and $\displaystyle y$.

3. Originally Posted by Nightasylum
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.
$\displaystyle \sqrt x + \sqrt y = 16 \Leftrightarrow \sqrt y = 16 - \sqrt x \Rightarrow$

$\displaystyle \Rightarrow \frac{d} {{dx}}\sqrt y = \frac{d} {{dx}}\left( {16 - \sqrt x } \right) \Leftrightarrow \frac{{dy}} {{dx}} \cdot \frac{1} {{2\sqrt y }} = - \frac{1} {{2\sqrt x }} \Leftrightarrow$

$\displaystyle \Leftrightarrow \frac{{dy}} {{dx}} = - \frac{{\sqrt y }} {{\sqrt x }} = - \frac{{16 - \sqrt x }} {{\sqrt x }} = 1 - \frac{{16}} {{\sqrt x }}.$