Question

what is the implicit derivate dy/dx about this expression:

(bx)^2 +(ay)^2 = (ab)^2

please folks!! help me

Answer 1

@zepdrix

Answer 2

@mathstudent55

Answer 3

\[2 bx*b+2ay*a\frac{ dy }{ dx }=0\]
\[2a^2y\frac{ dy }{ dx }=-2b^2x\]
\[\frac{ dy }{ dx }=-\frac{ b^2x }{ a^2y }\]

Answer 4

what about \[d^{2}y/dx^2\]

Answer 5

@zepdrix @Zeronknight @ShadowLegendX

Answer 6

Hold onto your first derivative, just keep it in your back pocket for now.\[\large\rm \color{indianred}{y'=\frac{-b^2x}{a^2y}}\]We'll need it later.

From here, how bout applying your quotient rule, ya?\[\large\rm y''=\left(\frac{-b^2x}{a^2y}\right)'\]

Answer 7

\[\large\rm y''=\frac{(-b^2x)'(a^2y)-(b^2x)(a^2y)'}{(a^2y)^2}\]