How do you find the derivative of #f(x) = 1/sqrt(2x-1)# by first principles?

Question

Isabella Ackerman · Accepted Answer

derivative of function to a power #-1/2 2(2x-1)^(-1/2-1)#

#f(x) = (2x-1)# You want the derivative of f to the power #-1/2# Chain rule #-1/2 2(2x-1)^(-1/2-1)#

The factor 2 comes from the derivative of f itself

Charles Arocha · Answer

Use limit definition of derivative to find:

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

#f(x) = 1/sqrt(2x-1)# #f'(a) = lim_(h->0) (f(a+h) - f(a))/h# #=lim_(h->0) (1/sqrt(2a+2h-1) - 1/sqrt(2a-1))/h# #=lim_(h->0) (sqrt(2a-1)-sqrt(2a+2h-1))/(h sqrt(2a+2h-1) sqrt(2a-1))# #=lim_(h->0) ((sqrt(2a-1)-sqrt(2a+2h-1))(sqrt(2a-1)+sqrt(2a+2h-1)))/(h sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))# #=lim_(h->0) ((2a-1)-(2a+2h-1))/(h sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))# #=lim_(h->0) (-2h)/(h sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))# #=lim_(h->0) (-2)/(sqrt(2a+2h-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a+2h-1)))# #=(-2)/(sqrt(2a-1) sqrt(2a-1) (sqrt(2a-1)+sqrt(2a-1)))# #=(-1)/((2a-1)^(3/2))# #=-(2a-1)^(-3/2)# So: #f'(x) = -(2x-1)^(-3/2)#

Charles Anctil · Answer

undefined To find the derivative of $ f(x) = \frac{1}{\sqrt{2x - 1}} $ using first principles, we start with the definition of the derivative:

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

First, we need to find $ f(x + h) $:

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

Now, substitute $ f(x + h) $ and $ f(x) $ into the difference quotient:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2(x + h) - 1}} - \frac{1}{\sqrt{2x - 1}}}{h} $$

Next, we rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{2x - 1} - \sqrt{2(x + h) - 1}}{h\sqrt{2(x + h) - 1}\sqrt{2x - 1}} $$

Now, combine the terms in the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{2x - 1} - \sqrt{2x + 2h - 1}}{h\sqrt{2(x + h) - 1}\sqrt{2x - 1}} $$

To simplify further, factor out $ -1 $ from the square root:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{2x - 1} - \sqrt{2x - 1 - 2h}}{h\sqrt{2(x + h) - 1}\sqrt{2x - 1}} $$

Now, we can cancel out the $ h $ terms:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{2x - 1} - \sqrt{2x - 1 - 2h}}{h} \cdot \frac{1}{\sqrt{2(x + h) - 1}\sqrt{2x - 1}} $$

Finally, take the limit as $ h $ approaches 0:

$$ f'(x) = \frac{-1}{2(x - 1)^{3/2}} $$

Therefore, the derivative of $ f(x) = \frac{1}{\sqrt{2x - 1}} $ by first principles is $ f'(x) = \frac{-1}{2(x - 1)^{3/2}} $.