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OpenStudy (anonymous):
If y^2-2xy=21, then dy/dx at the point (2,-3) is
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OpenStudy (anonymous):
implicit differentiation
2yy'+0xy-2y-2xy'=0
2yy'-2xy'=2y
yy'-xy'=y
y'(y-x)=y
y'=y/(y-x)
OpenStudy (anonymous):
plug in your values
OpenStudy (anonymous):
can u explain how u do implicit diff. ?
OpenStudy (anonymous):
im kinda bad at it
OpenStudy (anonymous):
use the chain rule whenever you take the derivative of y with respect to x:
for example, dy/dx (y^3)=3y^2y'
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OpenStudy (anonymous):
so when u differentiate, u put y' to show it was differentiated?
OpenStudy (anonymous):
dy/dx y=y'
OpenStudy (anonymous):
the chain rule is u multiply derivative of inside* outside + inside*derivative of outside, right?
OpenStudy (anonymous):
think you are talking about the product rule
OpenStudy (anonymous):
oh sorry so how do u use the chain rule?
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OpenStudy (anonymous):
so how did u know to use teh chain rule becuase here it isnt a compostion of funtions
OpenStudy (anonymous):
think \(y=f(x)\)
OpenStudy (anonymous):
\[y^2-2xy=21\]
\[f^2(x)-2xf(x)=521\]
OpenStudy (anonymous):
\[2f(x)f'(x)-2f(x)-2xf'(x)=0\] more easily written as
\[2yy'-2y-2xy'=0\]
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OpenStudy (anonymous):
ok i see that
OpenStudy (anonymous):
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