How do you find the derivative of #y = sqrt(x + 1)# using the limit definition?

Question

How do you find the derivative of #y = sqrt(x  + 1)# using the limit definition?

Emily Ammerman · Accepted Answer

We use the formula #f'(x) = lim_(h->0) (f(x+ h) - f(x))/h# to find the derivative.

#f'(x) = lim_(h->0) (sqrt(x + h +1) - sqrt(x + 1))/h#

Multiply this by the conjugate of the numerator, which is #sqrt(x+ h+ 1) + sqrt(x + 1)#.

#f'(x) = lim_(h->0) (sqrt(x + h + 1) - sqrt(x + 1))/h * (sqrt(x + h + 1) + sqrt(x + 1))/(sqrt(x + h + 1) + sqrt(x + 1))#

#f'(x) = lim_(h-> 0) (x+ h + 1 - (x + 1))/(hsqrt(x + h +1) + hsqrt(x + 1))#

#f'(x) = lim_(h->0) (x + h + 1- x - 1)/(h(sqrt(x + h + 1) + sqrt(x + 1)))#

#f'(x) = lim_(h->0) h/(h(sqrt(x + h + 1) + sqrt(x +1))#

#f'(x) = lim_(h->0) 1/(sqrt(x + h + 1) + sqrt(x + 1))#

You can now evaluate through substitution.

#f'(x) = 1/(sqrt(x + 0 + 1) + sqrt(x + 1))#

#f'(x) = 1/(2sqrt(x+ 1))#

Hopefully this helps!

Christopher Agresta · Answer

undefined To find the derivative of $ y = \sqrt{x + 1} $ using the limit definition, we start with the definition of the derivative:

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

Substitute $ f(x) = \sqrt{x + 1} $ into the definition:

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

To simplify, multiply the numerator and denominator by the conjugate of the numerator:

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

This simplifies to:

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

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

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

Now, as $ h $ approaches 0, the expression $ \sqrt{x + 1 + h} $ approaches $ \sqrt{x + 1} $, so:

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

Thus, the derivative of $ y = \sqrt{x + 1} $ with respect to $ x $ using the limit definition is:

$$ \frac{1}{2 \sqrt{x + 1}} $$