Question

find dy/dx at point (1,2) of the implicit function (x^2+y^2)^2=25/4xy^2.... my answer is -11/14.. am i right ? :))

Answer 1

Show your work.

Answer 2

\[2(x^2+y^2)(2x+2yy')=\frac{ 25 }{ 4 }.y^2+2yy'.\frac{ 25 }{ 4 }x\]

Answer 3

then i subtitute (1,2) to (x,y)

Answer 4

That would be correct then.

Answer 5

can u pls check it..................... :)

Answer 6

Ummm, just show the work... I won't do it but I'll check it.

Answer 7

\[2(1)^2+2(2)^2(2(1)+2(2).y'=\frac{ 25 }{ 4 }(2)+2(2).y'.\frac{ 25 }{ 4 }(1)\]

\[2+8(2+4y')=\frac{ 25 }{ 2 }+25y'\]

\[2+16+32y'=\frac{ 25 }{ 2 }+25y'\]

\[32y'-25y'=\frac{ 25 }{ 2 }-18\]

y'=\[y'=-\frac{ 11 }{1 4 }\]

Answer 8

well , what about this ?

Answer 9

Wow in the first step you get rid of important parenthesis.

Answer 10

huh ????? what ?

Answer 11

i dont get it

Answer 12

\[\begin{array}{rcl}
2(x^2+y^2)(2x+2yy')&=&\frac{ 25 }{ 4 }\cdot y^2+2yy'\cdot\frac{ 25 }{ 4 }x\\
2((1)^2+(2)^2)(2(1)+2(2)y')&=&\frac{ 25 }{ 4 }\cdot (2)^2+2(2)y'\cdot \frac{ 25 }{ 4 }(1)\\

2(1+4)(2+4y')&=&25 +y'\cdot 25\\
2(5)(2+4y')&=&25 +25y'\\
20+40y'&=&25 +25y'\\
15y'&=& 5\\
y'&=&\frac{1}{3}
\end{array}
\]How is this?

Answer 13

yeahh i miss that part , then the whole thing went worst..

Answer 14

and wrong...

Answer 15

@wio thanks a lot !!!!!!!!!! really appreciate it !!!!!!!! ^_^ i got it now :))

Answer 16

not best response?

Answer 17

of course best response ! :)