How do you find the derivative of #F(x)=arcsin sqrt sinx#?
↳Redirected from
"How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
Use the chain rule to find the derivative, because your function #F(x)# can be expressed as #G(h(j(x)))# where #G(x) = arcsin(x)#, #h(x) = x^(1/2)# and #j(x) = sin(x)#
The chain rule states that the derivative of a composite function is equal to the derivative of the individual composite functions with respect to their inner functions multiplied by the inner functions derivative with respect to x.
So #F'(x) = G'(h(j(x)) * h'(j(x)) * j'(x)#
Since #G'(x) = (1/(sqrt(1-x^2))), G'(h(j(x))) = 1/(sqrt(1 - (sqrt(sin(x)))^2)#
And #h'(x) = 1/2sqrt(x) , h'(j(x)) = 1/2(sqrt(sin(x)))#
And #j'(x) = cos(x)#
Thus the derivative becomes #(1/(sqrt(1 - (sqrt(sin(x)))^2))) * (1/(2(sqrt(sin(x))))) * (cos(x))#
Simplifying, we get #(cos(x))/(2*(sqrt(1-sin(x))) * (sqrt(sin(x))))#
This answer can be further simplified using trigonometric identities.
# F'(x) = cosx/(2 sqrtsinx sqrt(1-sinx) #
#F(x)=arcsin sqrt sinx#
Take sines of both sides:
#sin F(x)= sqrt sinx#
Implicit differentiation wrt x#:
#cos F(x) cdot F'(x) = 1/2 1/sqrtsinx cosx#
Re-arrange:
# F'(x) =cosx/(2 sqrtsinx ) cdot 1/ (cos F(x))#
# = cosx/(2 sqrtsinx) cdot 1/ (cos (arcsin sqrt sinx))#
Now draw a right-angled triangle and you will see that:
#cos ( arcsin theta) = sqrt(1-theta^2)#, or # arcsin theta = arccos( sqrt(1-theta^2))#
#implies F'(x) = cosx/(2 sqrtsinx sqrt(1-sinx) #
