Question

Derivative of y= x/square root of 1+x^2

Answer 1

This should remind you of the pythagorean theorem.

Answer 2

Try drawing the appropriate right triangle.

Answer 3

This has nothing to do with pythagorean theorem, it's calculus (Derivatives)

Answer 4

Oh, then you can completely ignore this then:


I'm sure \(\large \frac{x}{\sqrt{1+x^2}}\) couldn't possibly have anything to do with the sine function.

Answer 5

using quotient rule
y' = \[\sqrt{1+x^2} - 2x^2(\sqrt{1+x^2} / 1+x^2\]

Answer 6

the whole thing is divided by 1+x^2

and the quotient rule is this
[f'(x)g(x) - f(x)g'(x)] / g(x)^2

Answer 7

Or . . .
y = sin(Θ)
y' = cos(Θ)

Answer 8

See what I'm getting at here @Mcisn3 , @jayz657 ?

Answer 9

quotient rule for this is easy enough so that you do not need a trig sub to do this

Answer 10

Thanks jayz657, that makes sence :)