How do you find f'(x) using the definition of a derivative for #f(x)=1/sqrt(x)#?

Question

Christopher Agresta · Accepted Answer

The crucial step uses #(sqrtx - sqrt(x+h))(sqrtx + sqrt(x+h)) = -h#. If that's not enough of a hint, see below.

#f(x) = 1/sqrtx# #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h# # = lim_(hrarr0)(1/sqrt(x+h)-1/sqrtx)/h# # = lim_(hrarr0)((sqrtx-sqrt(x+h))/(sqrt(x+h)sqrtx))/(h/1)# # = lim_(hrarr0)((sqrtx-sqrt(x+h)))/((hsqrt(x+h)sqrtx)) # # = lim_(hrarr0)((sqrtx-sqrt(x+h)))/((hsqrt(x+h)sqrtx)) * ((sqrtx+sqrt(x+h)))/((sqrtx+sqrt(x+h)))# # = lim_(hrarr0)(x-(x+h))/((hsqrt(x+h)sqrtx)(sqrtx+sqrt(x+h)))# # = lim_(hrarr0)(-h)/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))# # = lim_(hrarr0)(-1)/(sqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))# # = (-1)/(sqrtxsqrtx(sqrtx+sqrtx))# # = (-1)/(2xsqrtx)# Or, if you prefer #(-1)/(2x^(3/2))#

Emilia Aguilar · Answer

undefined To find $ f'(x) $ using the definition of a derivative for $ f(x) = \frac{1}{\sqrt{x}} $, follow these steps:

1. Start with the definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the function $ f(x) = \frac{1}{\sqrt{x}} $ into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} $$

3. Rationalize the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h \sqrt{x} \sqrt{x + h}} $$

4. Combine the fractions in the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h} \times \frac{1}{\sqrt{x} \sqrt{x + h}} $$

5. Multiply and divide by the conjugate of the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h} \times \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}} $$

6. Simplify the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{x - (x + h)}{h(\sqrt{x} + \sqrt{x + h})} $$

7. Cancel out $ x $ terms in the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{-h}{h(\sqrt{x} + \sqrt{x + h})} $$

8. Cancel out $ h $ terms in the numerator and denominator:
$$ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x} + \sqrt{x + h}} $$

9. Substitute $ h = 0 $ into the expression:
$$ f'(x) = \frac{-1}{2\sqrt{x} \cdot 1} $$

10. Simplify the expression:
$$ \boxed{f'(x) = -\frac{1}{2x\sqrt{x}}} $$