My Math Forum mathematics - Q 8

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 February 8th, 2013, 06:15 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 mathematics - Q 8 Hi can please check my answer for a and b
 February 8th, 2013, 06:50 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: mathematics - Q 8 (B) is correct. For (A) you have $x^2+ y^2- 2cy+ c^2= 0$ $y'= 2x+ 2yy'- 2cy#39;= 0$ Surely, you understand that y' on the left is wrong? From $x^2+ y^2- 2cy+ c^2= 0$ you get $2x+ 2yy'- 2cy'= 0$ but NOT y'= 0. Differentiating again, $2+ 2(y'^2- 2yy''- 2cy''= 0" />. Now, going back to $yx+ 2yy'- 2cy'= 0$ ou can get $2cy'= yx+ 2yy#39;$ so that $c= \frac{yx+ 2yy'}{2y#39;}$. Replace the "c" in $yx+ 2yy'- 2cy'= 0$ with $\frac{yx+ 2yy'}{2y'}$.
 February 8th, 2013, 10:19 AM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: mathematics - Q 8 can explain to me line number 4 and 5 from where you got 2(y)^2 in line 4
 February 9th, 2013, 02:39 PM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: mathematics - Q 8 It's the product rule. We start with $2x+ 2yy'- 2xy'$ The derivative of 2x, with respect to x, is, of course, 2. The derivative of 2yy' is $2(y'(y'+ 2(y)(y''= 2y'^2+ 2yy''" />. The derivative of 2xy' is $2(x'y'+ 2x(y''= 2y'+ 2xy''" />. Putting those together, the derivative of $2x+ 2yy- 2xy'$ is $2+ 2y'^2+ 2yy''+ 2y'+ 2xy''$.

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