Question

Use the definition of the derivative as a limit to find the derivative f′ where f(x)= √ x+2.

Answer 1

Step-by-step explanation:

If the equation is

$\sqrt{x + 2}$

Then, here is the answer.

The definition of a derivative is

$\frac{f(x + h) - f(x)}{h}$

Also note that we want h to be a small, negligible value so we let h be a value that is infinitesimal small.

So we get

$\frac{ \sqrt{x + h + 2} - \sqrt{x + 2} }{h}$

Multiply both equations by the conjugate.

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

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

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

Since h is very small, get rid of h.

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

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

So the derivative of

$\frac{d}{dx} ( \sqrt{x + 2} ) = \frac{1}{2 \sqrt{x + 2} }$

Part 2: If your function is

$\sqrt{x} + 2$

Then we get

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

$\frac{ \sqrt{x + h} - \sqrt{x} }{h}$

$\frac{x + h - x}{h( \sqrt{x + h} + \sqrt{x}) }$

$\frac{h}{h( \sqrt{x + h} + \sqrt{x} ) }$

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

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

So

$\frac{d}{dx} ( \sqrt{x} + 2) = \frac{1}{2 \sqrt{x} }$