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)#

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