How do you use the definition of a derivative to find the derivative of #1/sqrt(x)#?

Question

Isabella Ackerman · Accepted Answer

First, remember that square roots can be rewritten in exponential forms:

#root(n)(x^m)# = #x^(m/n)#

As you have a simple square root in the denominator of your function, we can rewrite it as #x^(1/2)#, alright?

Now, remembering that when potences are on the denominator you can bring them to the numerator by changing its positivity/negativity, you can rewrite #1/(x^(1/2))# as #x^(-1/2)#

Deriving now your function, you'll get:

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

Alternatively: #-1/(2(x^(3/2))#

Or even in this form: #-1/(root(2)(x^3))#

Ezra Adams · Answer

#f(x)=1/sqrtx# #f'(x)=lim_(hrarr0) (f(x+h)-f(x))/h# # = lim_(hrarr0) (1/sqrt(x+h)-1/sqrt(x))/h# # = lim_(hrarr0) ((sqrtx-sqrt(x+h))/(sqrt(x+h)sqrt(x)))/(h/1)# # = lim_(hrarr0) (sqrtx-sqrt(x+h))/(sqrt(x+h)sqrt(x)) * 1/h# # = lim_(hrarr0) ((sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrt(x)) * ((sqrtx+sqrt(x+h)))/((sqrtx+sqrt(x+h)))# # = lim_(hrarr0) (x-(x+h))/(hsqrt(x+h)sqrt(x)(sqrtx+sqrt(x+h)))# # = lim_(hrarr0) (-h)/(hsqrt(x+h)sqrt(x)(sqrtx+sqrt(x+h)))# # = lim_(hrarr0) (-1)/(sqrt(x+h)sqrt(x)(sqrtx+sqrt(x+h)))# # =  (-1)/(sqrt(x+0)sqrt(x)(sqrtx+sqrt(x+0)))# # =  (-1)/(x(2sqrtx)) =(-1)/(2xsqrtx) = (-1)/(2x^(3/2))#

Aurora Alvarado · Answer

undefined To find the derivative of $ \frac{1}{\sqrt{x}} $ using the definition of a derivative, we start with the definition:

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

Substitute $ f(x) = \frac{1}{\sqrt{x}} $:

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

Multiply the numerator and denominator by $ \sqrt{x + h} \sqrt{x} $ to rationalize the expression:

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

Now, simplify the expression:

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

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

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

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

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

Therefore, the derivative of $ \frac{1}{\sqrt{x}} $ with respect to $ x $ is $ -\frac{1}{2x \sqrt{x}} $.