Implicit Differentiation

nychic

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Jun 22, 2012
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Hey I seem to be getting this question wrong, please help thanks :):

Find the derivative of the function y defined implicitly in terms of x. y =
sqrt1a.gif
xy+ 9


 
Hey I seem to be getting this question wrong, please help thanks :):

Find the derivative of the function y defined implicitly in terms of x. y =
sqrt1a.gif
xy+ 9

Show us what you have done so far so we can know where you are getting stuck. Thanks!
 
y=sqrt(xy+9)
= (xy+9)^1/2
y'=1/2(xy+9)^-1/2 * d/dx(xy+9)
=1/2(xy+9)^1/2* (x)(dy/dx)+(y)(1)

I know I can put the (xy+9)^-1/2 down to the denominator to make it positive but I'm not sure if I use the product rule when taking the derivative from the inside function (xy+9). It doesn't seem that dy/dx should be included in the numerator of this problem. So I don't know what to do next.
 
y=sqrt(xy+9)
= (xy+9)^1/2
y'=1/2(xy+9)^-1/2 * d/dx(xy+9)
=1/2(xy+9)^1/2* (x)(dy/dx)+(y)(1)

I know I can put the (xy+9)^-1/2 down to the denominator to make it positive but I'm not sure if I use the product rule when taking the derivative from the inside function (xy+9). It doesn't seem that dy/dx should be included in the numerator of this problem. So I don't know what to do next.

Good start, although it is a bit hard to follow without proper parentheses notation. Half the battle with implicit differentiation is the simplification of the data to solve for y'.

\(\displaystyle y=\sqrt{xy+9} = (xy+9)^{1/2}\)

\(\displaystyle y'=[\frac{1}{2}(xy+9)^{-1/2}](y+xy')=\frac{y+xy'}{2\sqrt{xy+9}}\)


cross multiply and get:

\(\displaystyle 2y'\sqrt{xy+9}=y+xy'\)

\(\displaystyle 2y'\sqrt{xy+9}-xy'=y\)

\(\displaystyle y'(2\sqrt{xy+9}-x)=y\)

\(\displaystyle y'=\frac{y}{2\sqrt{xy+9}-x}\)
 
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