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

Question

Aurora Alvarado · Accepted Answer

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]"#

Scarlett Allison · Answer

undefined To find the derivative of $ \frac{x}{\sqrt{1-x^2}} $ using the limit definition of the derivative, we first express the function in a form suitable for this method.

The limit definition of the derivative of a function $ f(x) $ is given by:

$$ f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} $$

In this case, let $ f(x) = \frac{x}{\sqrt{1-x^2}} $. To find $ f'(x) $, we need to evaluate the limit as $ h \to 0 $ of the expression $ \frac{f(x + h) - f(x)}{h} $.

First, find $ f(x + h) $:

$$ f(x + h) = \frac{x + h}{\sqrt{1 - (x + h)^2}} $$

Next, find $ f(x) $:

$$ f(x) = \frac{x}{\sqrt{1 - x^2}} $$

Now, subtract $ f(x) $ from $ f(x + h) $:

$$ f(x + h) - f(x) = \frac{x + h}{\sqrt{1 - (x + h)^2}} - \frac{x}{\sqrt{1 - x^2}} $$

Simplify the expression:

$$ f(x + h) - f(x) = \frac{(x + h)\sqrt{1 - x^2} - x\sqrt{1 - (x + h)^2}}{\sqrt{1 - x^2}\sqrt{1 - (x + h)^2}} $$

Now, divide by $ h $ and take the limit as $ h \to 0 $ to find the derivative. This process involves some algebraic manipulation and can be complex, so it's often easier to use trigonometric substitution or other methods to simplify the expression before taking the limit.