What is the derivative of # arcsin(1/x)#?

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Answer 1 · 2015-07-26T20:43:19

#-1/(xsqrt(x^2-1))#

To differentiate this we will be applying a chain rule:

Start by Letting #theta=arcsin(1/x)#

#=>sin(theta)=1/x#

Now differentiate each term on both sides of the equation with respect to #x#

#=>cos(theta)*(d(theta))/(dx)=-1/x^2#

Using the identity : #cos^2theta+sin^2theta=1 => costheta=sqrt(1-sin^2theta)#

#=>sqrt(1-sin^2theta)*(d(theta))/(dx)=-1/x^2#

#=>(d(theta))/(dx)=-1/x^2*1/sqrt(1-sin^2theta)#

Recall : #sin(theta)=1/x" "# and #" "theta=arcsin(1/x)#

So we can write,

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

#=-1/x^2*x/sqrt(x^2-1)=color(blue)(-1/(xsqrt(x^2-1)))" or "-sqrt(x^2-1)/(x(x^2-1))#