What is the derivative of #1/sqrt(1 - x^2)#?

2 Answers
Apr 12, 2018

#=(x)(1-x^2)^(-3/2)#

Explanation:

#f(x)=1/sqrt(1-x^2)=(sqrt(1-x^2))^-1=(1-x^2)^(-1/2)#
#d/dx f(x)=-1/2(1-x^2)^(-1/2-1)(0-2x)#
#=(2x)/2(1-x^2)^(-3/2)#
#=(x)(1-x^2)^(-3/2)#

Apr 12, 2018

The derivative is #(xsqrt(1-x^2))/(1-x^2)^2#.

Explanation:

Using the quotient rule:

#color(white)=d/dx[1/sqrt(1-x^2)]#

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

#=(0*sqrt(1-x^2)-1*d/dx[sqrt(1-x^2)])/(1-x^2)#

#=(-d/dx[sqrt(1-x^2)])/(1-x^2)#

Chain rule:

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

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

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

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

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

That's the derivative. Hope this helped!