If #f(x)= sqrt(x^2-1 # and #g(x) = 1/x #, what is #f'(g(x)) #?

Question

If #f(x)=  sqrt(x^2-1 # and #g(x)  = 1/x  #, what is #f'(g(x))  #?

Elijah Abram · Accepted Answer

#f'(g(x)) = 1/sqrt(1-x^2)# #f(x) = sqrt(x^2-1) = (x^2-1)^(1/2)# #f'(x) = 1/2(x^2-1)^(-1/2)*2x# (Power rule and Chain rule) #= x/sqrt(x^2-1)# Given #g(x) = 1/x# #f'(g(x)) = (1/x)/sqrt((1/x)^2-1)# #= 1/(x(sqrt((1-x^2)/x^2))# #=1/(x/x*sqrt((1-x^2))# #= 1/(sqrt((1-x^2))#

Isabella Ackerman · Answer

undefined To find $ f'(g(x)) $, you first need to find the derivative of $ f(x) $ with respect to $ x $, and then substitute $ g(x) $ for $ x $ in the resulting expression.

1. Derivative of $ f(x) = \sqrt{x^2 - 1} $ with respect to $ x $:
$$ f'(x) = \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}} $$

2. Substitute $ g(x) = \frac{1}{x} $ into $ f'(x) $:
$$ f'(g(x)) = \frac{g(x)}{\sqrt{g(x)^2 - 1}} = \frac{\frac{1}{x}}{\sqrt{\left(\frac{1}{x}\right)^2 - 1}} $$

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

$$ f'(g(x)) = \frac{\frac{1}{x}}{\frac{\sqrt{1 - x^2}}{|x|}} = \frac{1}{x} \cdot \frac{|x|}{\sqrt{1 - x^2}} $$

$$ \therefore f'(g(x)) = \frac{|x|}{x\sqrt{1 - x^2}} $$

Note: Since $ g(x) = \frac{1}{x} $, $ x $ cannot be zero, hence $ x \neq 0 $. Thus, the absolute value $ |x| $ is redundant. Therefore, the final expression can be simplified to:

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