Question

How do you find the derivative of `y = sqrt(1-sin^2x)`?

Answer 1

`y' = -sinx`

Explanation:

By the pythagorean identity `sin^2x + cos^2x = 1`:

`y = sqrt(cos^2x)`

`y= cosx`

This can be differentiated as `-sinx`, but here's proof using limits.

`y' = lim_(h->0) (f(x + h)- f(x))/h`

`y' = lim_(h->0) (cos(x+ h) - cosx)/h`

`y' = lim_(h->0) (cosxcos h - sinxsin h - cosx)/h`

`y' = lim_(h->0) (cosxcos h - cosx)/h - (sinxsin h)/h`

`y' = lim_(h-> 0) (cosx(cos h - 1))/h - lim_(h->0) (sinxsin h)/h`

`y' = cosxlim_(h->0)(cos h - 1)/h - sinxlim_(h->0) (sin h)/h`

Now use the famous trig limits:

`•lim_(x->0) (cosx - 1)/x = 0`
`•lim_(x->0) sinx/x = 1`

`y' = cosx(0) - sinx(1)`

`y' = -sinx`

Hopefully this helps!