Another Implicit Diff Problem

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Jason76

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Post Edited 9/27/13

Right??

\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0\) - Product Rule on \(\displaystyle 3xy\)

\(\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + 3y + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + 3y + y'(3x - 2y) = 0\)

\(\displaystyle y'(3x - 2y) = -16x - 3y\)

\(\displaystyle y'= \dfrac{-16x - 3y}{3x - 2y}\) :confused:
 
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Jason76 said:
Solving for y'
\(\displaystyle 8x^{2} + 3xy - y^{2} = 2\)
Click to expand...

Do everything nice and slowly.
Note how I got rid of unnecessary symbols and grouped the derivatives together.

\(\displaystyle \\(16x)+(3y+3xy')-(2yy')=0\\(3x-2y)y'=-16x-3y\\y'=-\dfrac{16x+3y}{3x-2y}\)
 
Jason76 said:
Solving for y'

\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle f'(x) = 16x + 3yy' - 2yy' = 0\)
Click to expand...
You have made the same mistake you did with the other one- not using the "product rule".
The derivative (uv)' is u'v+ uv'. The derivative of 3xy, with respect to x, is (3x)'y+ 3x(y)'= 3y+ 3xy'.

\(\displaystyle f'(x) = 3yy' - 2yy' = -16x\)

\(\displaystyle f'(x) = yy'(3 - 2) = -16x\)

\(\displaystyle f'(x) = yy'(3 -2)(\dfrac{1}{3 - 2}) = -16x (\dfrac{1}{3 -2})\)

\(\displaystyle f'(x) = yy' = -\dfrac{16x}{3 - 2}\)

\(\displaystyle f'(x) = yy'(\dfrac{1}{y}) = -\dfrac{16x}{3 - 2}(\dfrac{1}{y})\)

\(\displaystyle f'(x) = y' = -\dfrac{16x}{3 - 2y}\):confused:
Click to expand...
 
HallsofIvy said:
You have made the same mistake you did with the other one- not using the "product rule".
The derivative (uv)' is u'v+ uv'. The derivative of 3xy, with respect to x, is (3x)'y+ 3x(y)'= 3y+ 3xy'.
Click to expand...

I see the mistake now, not using the product rule.
 
Right??

\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0\) - Product Rule on \(\displaystyle 3xy\)

\(\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + 3y + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + 3y + y'(3x - 2y) = 0\)

\(\displaystyle y'(3x - 2y) = -16x - 3y\)

\(\displaystyle y'= \dfrac{-16x - 3y}{3y - 2y}\)
 
Last edited:
Jason76 said:
Right??

\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3)y' + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + (3)(y)y' + (3x)(1) - 2yy' = 0\)

\(\displaystyle 16x + 3yy' + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + y'(3y + 3x - 2y) = 0\)

\(\displaystyle y'(3y + 3x - 2y) = -16x\)

\(\displaystyle y' = \dfrac{16x}{3y + 3x - 2y}\)
Click to expand...

Look at reply #2.
 
pka says to do one thing, while Halls of Ivy says to use the product rule. Which do I do?
 
Jason76 said:
pka says to do one thing, while Halls of Ivy says to use the product rule. Which do I do?
Click to expand...

Look at pka's response very carefully - pka is using product rule (along with power rule) - you cannot escape that!!
 
\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0\) - Product Rule on \(\displaystyle 3xy\)

\(\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + 3y + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + 3y + y'(3x - 2y) = 0\)

\(\displaystyle y'(3x - 2y) = -16x - 3y\)

\(\displaystyle y'= \dfrac{-16x - 3y}{3y - 2y}\) :confused:
 
The online homework is still saying this is wrong. Did I not simpify it enough?
 
Jason76 said:
\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0\) - Product Rule on \(\displaystyle 3xy\)

\(\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + 3y + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + 3y + y'(3x - 2y) = 0\)

\(\displaystyle y'(3x - 2y) = -16x - 3y\)

\(\displaystyle y'= \dfrac{-16x - 3y}{3y - 2y}\) ...........................This is incorrect


:confused:
Click to expand...
.
 
Subhotosh Khan said:
.
Click to expand...


\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0\) - Product Rule on \(\displaystyle 3xy\)

\(\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + 3y + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + 3y + y'(3x - 2y) = 0\)

\(\displaystyle y'(3x - 2y) = -16x - 3y\)

\(\displaystyle y'= \dfrac{-16x - 3y}{3x - 2y}\)

This answer is correct on online homework, but wouldn't the x's and y's cancel out?
 
Jason76 said:
\(\displaystyle f(x) = 8x^{2} + 3xy - y^{2} = 2\)

\(\displaystyle 16x + (y)(3) + (3x)(1)y' - 2yy' = 0\) - Product Rule on \(\displaystyle 3xy\)

\(\displaystyle 16x + (3)(y) + (3x)(1)y' - 2yy' = 0\)

\(\displaystyle 16x + 3y + 3xy' - 2yy' = 0\)

\(\displaystyle 16x + 3y + y'(3x - 2y) = 0\)

\(\displaystyle y'(3x - 2y) = -16x - 3y\)

\(\displaystyle y'= \dfrac{-16x - 3y}{3x - 2y}\)

This answer is correct on online homework, but wouldn't the x's and y's cancel out?
Click to expand...

No - what exactly do you mean by "cancel out"?
 
Jason76 said:
\(\displaystyle y'= \dfrac{-16x - 3y}{3x - 2y}\)

...wouldn't the x's and y's cancel out?
Click to expand...
:shock:

No.

Seriously, you really need to learn algebra before you attempt calculus. To learn what it means to "simplify a rational expression" (and about "cancelling" in particular), try here. ;)
 
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