How do you find the derivative of y = sqrt(1-sin^2x)y=√1−sin2x?
1 Answer
Jan 22, 2017
Explanation:
By the pythagorean identity
y = sqrt(cos^2x)
y= cosx
This can be differentiated as
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!
