How do you find the derivative of #arccos(x^2)#?

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Answer 1 · 2015-02-20T19:28:17

The answer is: #y'=-(2x)/sqrt(1-x^4)#

Since #y=arccosf(x)rArry'=-f(x)/sqrt(1-[f(x)]^2#,

than:

#y'=-(2x)/sqrt(1-x^4)#.

Answer 2 · 2015-02-21T21:35:57

We have

#y=arccos(x^2) #

Take the cosine of both sides

#cos(y)=cos(arccos(x^2)) #

#cos(y)=x^2 #

Now differentiate both sides with respect to #x#

#-sin(y)dy/dx=2x #

Divide both sides by #-sin(y)#

#dy/dx=(2x)/-sin(y) #

Recall that #sin^2y+cos^2y=1# Therefore,

#sin^2y=1-cos^2y #

#sin(y)=sqrt(1-cos^2y)# we restrict ourselves to the positive root

From above #cos(y)=x^2#. Consequently

#cos^2y=x^4 #

So we can write

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