Question

Find the second derivative of 3x^2-5y^2=1.

Answer 1

I have the first derivative. \[\frac{ dy }{ dx } = \frac{ 6x }{ 10y }\]

Answer 2

(3*2)x^2-1 - 2*5y^2-1(dy/dx)=0
\[6x -10y \frac{ dy }{ dx }=0\]\[\frac{ 6x }{ 10y }=\frac{ dy }{ dx }\]\[\frac{ 3x }{ 5y }=\frac{ dy }{ dx }\]\[\frac{ (5y)(3)-(3x)(5\frac{ dy }{ dx }) }{ (5y)^{2} }\]\[\frac{ 15y-15x(\frac{ 3x }{ 5y}) }{ (5y)^2 }\]....can u simplify frm here...?

Answer 3

i used the quotient rule...

Answer 4

I can get that far with the quotient rule and understand why. Simplification is where I start to get careless I think.....

Answer 5

\[\frac{ 15y-3x(\frac{ 3x }{ y }) }{ (5y)^2 }\]\[\frac{ 15y-\frac{ 9x^2 }{ y } }{ (5y)^2 }\]\[\frac{ \frac{ (15y^2-9x^2) }{ y } }{ (5y)^2}\]

Answer 6

\[\frac{ 15y^2-9x^2 }{ y }*\frac{ 1 }{ (5y)^2 }\]

Answer 7

\[\frac{ 15y^2-9x^2 }{ (y)(5y)^2 }\]\[\frac{ 15y^2-9x^2 }{ (y) (25y^2)}\]\[\frac{ 15y^2 }{ 25y^3 }-\frac{ 9x^2 }{ 25y^3 }\]

Answer 8

\[\frac{ 3 }{ 5y}-\frac{ 9x^2 }{ 25y^3 }\]

Answer 9

ok thats the simplest i can go with this