What is the derivative of sin(arccos x)?

3 Answers
maganbhai P.
Aug 14, 2018

(dy)/(dx)=-(cos(arc cosx))/sqrt(1-x^2)=-x/sqrt(1-x^2)

Explanation:

Here,

y=sin(arc cosx)

Let ,

y=sinu and u=arc cosx

:.(dy)/(du)=cosu and (du)/(dx)=-1/sqrt(1-x^2)

Using Chain Rule :

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)

:.(dy)/(dx)=cosu ((-1)/sqrt(1-x^2))=-cosu/sqrt(1-x^2)

Subst. back u=arc cosx

:.(dy)/(dx)=-(cos(arc cosx))/sqrt(1-x^2)=-x/sqrt(1-x^2)

maganbhai P.
Aug 14, 2018

(dy)/(dx)=-x/sqrt(1-x^2)

Explanation:

Here,

y=sin(arc cosx)

:.y=sin(arcsin(sqrt(1-x^2))

:.y=sqrt(1-x^2)

:.(dy)/(dx)=1/(2sqrt(1-x^2))d/(dx)(1-x^2)

:.(dy)/(dx)=1/(2sqrt(1-x^2))(-2x)

:.(dy)/(dx)=-x/sqrt(1-x^2)

George C.
Aug 14, 2018

d/(dx) sin(arccos(x)) = -x/sqrt(1-x^2)

Explanation:

Note that for x in [-1, 1] we have arccos(x) in [0, pi] and sin(arccos(x)) >= 0.

So using cos^2 theta + sin^2 theta = 1 we have:

sin(arccos(x)) = sqrt(1-x^2) = (1-x^2)^(1/2)

Hence:

d/(dx) sin(arccos(x)) = d/(dx) (1-x^2)^(1/2)

color(white)(d/(dx) sin(arccos(x))) = 1/2 (1-x^2)^(-1/2) * d/(dx) (1-x^2)

color(white)(d/(dx) sin(arccos(x))) = 1/2 (1-x^2)^(-1/2) * (-2x)

color(white)(d/(dx) sin(arccos(x))) = -x/sqrt(1-x^2)

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
12540 views around the world
You can reuse this answer
Creative Commons License