How do you use the definition of a derivative to find the derivative of #(2/sqrt x)#?

Question

Emily Ammerman · Accepted Answer

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

The limit definition of a derivative states that for a function #f(x)#, its derivative is

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

Here, #f(x)=2/sqrtx#, so

#f'(x)=lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h#

Multiply the entire fraction by #sqrt(x+h)sqrtx#.

#f'(x)=lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h((sqrt(x+h)sqrtx)/(sqrt(x+h)sqrtx))#

#f'(x)=lim_(hrarr0)(2sqrtx-2sqrt(x+h))/(hsqrt(x+h)sqrtx)#

#f'(x)=lim_(hrarr0)(2(sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrtx)#

Multiply by the conjugate of the term in the numerator.

#f'(x)=lim_(hrarr0)(2(sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrtx)((sqrtx+sqrt(x+h))/(sqrtx+sqrt(x+h)))#

#f'(x)=lim_(hrarr0)(2(x-(x-h)))/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#

#f'(x)=lim_(hrarr0)(2h)/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#

#f'(x)=lim_(hrarr0)2/(sqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#

Evaluate the limit by plugging in #0# for #h#.

#f'(x)=2/(sqrt(x+0)sqrtx(sqrtx+sqrt(x+0)))#

#f'(x)=2/(sqrtxsqrtx(sqrtx+sqrtx)#

#f'(x)=2/(x(2sqrtx))#

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

Christopher Allers · Answer

undefined To find the derivative of $ \frac{2}{\sqrt{x}} $ using the definition of a derivative, we'll first express it as a function of $ x $:

$$ f(x) = 2x^{-1/2} $$

Then, we'll use the definition of the derivative:

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

Substitute $ f(x) = 2x^{-1/2} $ into the definition:

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

Now, we'll simplify the expression:

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

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

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

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

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

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

Now, plug in $ h = 0 $ to find the derivative:

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

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

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