How do you find the derivative of #sin(arccosx)#?

2 Answers
Aug 1, 2015

Use the Chain Rule, resulting in #cos(cos^-1(x))*(-1)/sqrt(1-x^2)#

Explanation:

The Chain Rule involves breaking a function up into a function of a function, sometimes expressed as #y=f(u)# and #u=g(x)#.

Step 1. Let #y=sin(u)# and #u=cos^-1(x)#.

Step 2. Chain Rule says #dy/dx=(dy)/(du)*(du)/(dx)#.

Step 3. Apply the Chain Rule to the equations in Step 1.

#(dy)/(du)=cos(u)# and #(du)/(dx)=(-1)/(sqrt(1-x^2)#.

Step 4. Get rid of the #u#'s in #(dy)/(du)#. Recall that #u=cos^-1(x)#, so then

#(dy)/(du)=cos(cos^-1(x))#

Step 5. Plug your derivatives back into Chain Rule of Step 2.

#dy/dx=(dy)/(du)*(du)/(dx)=cos(cos^-1(x))*(-1)/(sqrt(1-x^2)#

Step 6 is often optional, but you can then try to simplify as much as you can.

#dy/dx=(dy)/(du)*(du)/(dx)=(-cos(cos^-1(x)))/(sqrt(1-x^2)#.

Aug 1, 2015

I would use the fact that #sin(arccosx) = sqrt(1-x^2)#

Explanation:

#arccosx# is an angle (or, a number) between #0# and #pi# whose cosine is #x#.
The sine of an angle (or, a number) between #0# and #pi# whose cosine is #x# is #sqrt(1-x^2)#.
(Use #sin^2 theta + cos^2 theta = 1#. So #cos theta = +-sqrt(1-sin^2 theta)#
and with #theta# in #[0, pi]#, we have #cos theta > 0#.)

#sin(arccosx) = sqrt(1-x^2)#

Now i'll use #d/dx(sqrtu) = 1/(2sqrtu) (du)/dx# to get:

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

# = 1/(2sqrt(1-x^2)) * (-2x)#

# = (-x)/sqrt(1-x^2)#