How do you use the limit definition to find the derivative of #x/(sqrt(1-x^2)#?

1 Answer
Jul 19, 2017

Given #f(x) = x/sqrt(1-x^2)" [1]"#

#f(x+h)= (x+h)/sqrt(1-(x+h)^2)" [2]"#

#f'(x) = lim_(hto0) (f(x+h)-f(x))/h" [3]"#

Substitute equation [1] and [2] into equation [3]:

#f'(x) = lim_(hto0) ((x+h)/sqrt(1-(x+h)^2)-x/sqrt(1-x^2))/h" [3.1]"#

Make a common denominator by multiplying the numerator and denominator by 1 in the form #(sqrt(1-(x+h)^2)sqrt(1-x^2))/(sqrt(1-(x+h)^2)sqrt(1-x^2))#:

#f'(x) = lim_(hto0) ((x+h)sqrt(1-x^2)-xsqrt(1-(x+h)^2))/(hsqrt(1-(x+h)^2)sqrt(1-x^2))" [3.2]"#

Multiply numerator and denominator by 1 in the form #((x+h)sqrt(1-x^2)+xsqrt(1-(x+h)^2))/((x+h)sqrt(1-x^2)+xsqrt(1-(x+h)^2))# this will make the numerator become the difference of two squares:

#f'(x) = lim_(hto0) ((x+h)^2(1-x^2)-x^2(1-(x+h)^2))/(hsqrt(1-(x+h)^2)sqrt(1-x^2)((x+h)sqrt(1-x^2)+xsqrt(1-(x+h)^2)))" [3.3]"#

Simplify the numerator (a lot):

#f'(x) = lim_(hto0) (h^2+2hx)/(hsqrt(1-(x+h)^2)sqrt(1-x^2)((x+h)sqrt(1-x^2)+xsqrt(1-(x+h)^2)))" [3.4]"#

There is common factor of #h/h to 1#:

#f'(x) = lim_(hto0) (h+2x)/(sqrt(1-(x+h)^2)sqrt(1-x^2)((x+h)sqrt(1-x^2)+xsqrt(1-(x+h)^2)))" [3.5]"#

We may, now, let #h to 0# without any problems:

#f'(x) = (2x)/(sqrt(1-(x)^2)sqrt(1-x^2)((x)sqrt(1-x^2)+xsqrt(1-(x)^2)))" [3.6]"#

Simplify the denominator:

#f'(x) = (2x)/((2x)(1-(x)^2)sqrt(1-x^2))" [3.7]"#

#(2x)/(2x) to 1 # and the denominator becomes the #3/2# power:

#f'(x) = 1/(1-x^2)^(3/2)" [3.8]"#