Question

FIND THE DERIVATIVE OF

y^2= b^2/2x+b at O,b

please help me.

Answer 1

Is it:

\[\large y^2 = \frac{b^2}{2x} + b\]

Answer 2

b is in the denominator with 2x

Answer 3

\[\large y^2 = \frac{b^2}{2x+b} \]

Answer 4

Take the derivative implicitly..

Answer 5

do what @waterineyes said
do you know this?

Answer 6

yes it is implicit differentiation.

Answer 7

yy'=-b^2/(2x+b)^2

Answer 8

Implicitly means:

\[2y.dy = (b^2(-1)(2x+b)^{-2} \times 2).dx\]


\[\large y.dy = (-b^2(2x + b)^{-2}).dx\]

\[\large \frac{dy}{dx} = \frac{-b^2}{y(2x + b)^2}\]

Now put x = 0 and y = b here..

Answer 9

thanks @waterineyes! ((=

Answer 10

Welcome dear..

Answer 11

@waterineyes wait where did you get -1?

Answer 12

Tell me the derivative of \(x^{-2}\) ??

Answer 13

-2x^-3

Answer 14

Yes I have done the same there..

Answer 15

I have written:

\[\frac{1}{2x+b} \quad as \quad (2x+b)^{-1}\]

Now take its derivative..

Getting??

Answer 16

oooh i get it. thanks a lot

Answer 17

\[\frac{d}{dx}(2x + b)^{-1} = (-1) \times (2x + b)^{-2} \times 2\]

Answer 18

Thanks! =]]

Answer 19

Welcome buddy..