Equation of tangent line to curve

**togo** · September 10th 2011, 01:35 PM

given:

$xy^\frac{1}{2} + yx^\frac{1}{2} = 6$

we are trying to find $\frac{dy}{dx}$ so then, wouldn't you want to isolate y?

eventually you want a $m_2=\frac{1}{m_1}$

instead the book starts turning the original formula on this page into a derivative. I am in grants notes and have been stuck on this question for some time. If I work on it more I'll update this but if anyone thinks they can explain it, please do.

**Plato** · September 10th 2011, 02:20 PM

Originally Posted by togo

given:
$xy^{\frac{1}{2}} + yx^{\frac{1}{2}} = 6$

we are trying to find $\frac{dy}{dx}$

$y^{\frac{1}{2}}+\frac{1}{2}xy^{-\frac{1}{2}}y'+x^{\frac{1}{2}}y'+\frac{1}{2}x^{-\frac{1}{2}}y=0$

Solve for $y'.$

**HallsofIvy** · September 10th 2011, 02:38 PM

Originally Posted by togo

given:

$xy^\frac{1}{2} + yx^\frac{1}{2} = 6$

we are trying to find $\frac{dy}{dx}$ so then, wouldn't you want to isolate y?

Plato used a method called "implicit differentiation" where he did not need to isolate y.

eventually you want a $m_2=\frac{1}{m_1}$

instead the book starts turning the original formula on this page into a derivative. I am in grants notes and have been stuck on this question for some time. If I work on it more I'll update this but if anyone thinks they can explain it, please do.

Yes, that's what "implicit differentiation" is. For example, differentiating " $xy^2$ " with respect to x, you first use the product rule to say that is [tex](x)'y^2+ x(y^2)'[tex]. Of course, the derivative of x, with respect to x, is 1 and the derivative of $y^2$ with respect to x is, by the chain rule, the derivative of $y^2$ with respect to y times the derivative with respect to x: 2yy'. The is, the derivative of $xy^2$ with respect to x is $(1)y^2+ x(2y)y'= y^2+ 2xyy'$ .

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