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

Question

Scarlett Allison · Accepted Answer

#y' = -sinx#

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!

Charles Anctil · Answer

undefined To find the derivative of $ y = \sqrt{1 - \sin^2(x)} $, you can use the chain rule and the fact that the derivative of $ \sqrt{u} $ with respect to $ u $ is $ \frac{1}{2\sqrt{u}} $.

So, $ \frac{dy}{dx} = \frac{1}{2\sqrt{1 - \sin^2(x)}} \cdot \frac{d}{dx}(1 - \sin^2(x)) $.

Using the chain rule, $ \frac{d}{dx}(1 - \sin^2(x)) = -2\sin(x)\cos(x) $.

Substitute this result back into the equation to find $ \frac{dy}{dx} $.

Thus, $ \frac{dy}{dx} = -\frac{\sin(x)\cos(x)}{\sqrt{1 - \sin^2(x)}} $.