What is the derivative of #g(x)=3 arccos (x/2)#?

Question

2087 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-12T16:11:23

The answer is #=-3/2*1/sqrt(1-x^2/4)#

We use,

#sin^2theta+cos^2theta=1#

Let #y=3arccos(x/2)#

Rearranging

#y/3=arccos(x/2)#

#cos(y/3)=x/2#

#cos^2(y/3)+sin^2(y/3)=1#

#sin^2(y/3)=1-cos^2(y/3)=1-x^2/4#

#sin(y/3)=sqrt(1-x^2/4)#

Differentiating, we get

#(cos(y/3))'=(x/2)'#

#-1/3sin(y/3)*dy/dx=1/2#

#dy/dx=-3/2*1/sin(y/3)#

#=-3/2*1/sqrt(1-x^2/4)#