Question

y=-1/√(x^2+1) then dy/dx=

Answer 1

Any ideas?

Answer 2

you can rewrite it as
\[y = (x^2+1)^{-1/2}\]
take dy/dx

Answer 3

How do you take dy/dx?

Answer 4

\[y' = -\frac{ 1 }{ 2 } (x^2+1)^{\frac{ 1 }{ 2 }}(2x)\]

Answer 5

Isn't that the derivative? I thought dy/dx was implicit differentiation?

Answer 6

\[dy= f'(x) \Delta x, dy= f'(x) dx\]

Answer 7

\[dx= \frac{ dy }{ f'(x) }\]

Answer 8

It is not implicit differentiation. If you have:
\[y=f(x)\]
Then the derivative can be written in several forms such as:
\[\frac{dy}{dx}=f'(x)=\frac{df}{dx}\]

Answer 9

Ohhhhhh... Thank you!