How do you find the derivative of $\sqrt{x + 1}$ $sqrt( x+1 )$ using limits?

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Answer 1 · 2017-03-21T13:06:12

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

By definition:

$(df)/dx = lim_(h->0) ( f(x+h)-f(x))/h$

For $f(x) = sqrt(x+1)$ we have:

$d/dx(sqrt(x+1)) = lim_(h->0)(sqrt(x+h+1)-sqrt(x+1))/h$

Multiply and divide the function by $(sqrt(x+h+1)+sqrt(x+1))$ :

$d/dx(sqrt(x+1)) = lim_(h->0)((sqrt(x+h+1)-sqrt(x+1))/h)((sqrt(x+h+1)+sqrt(x+1))/(sqrt(x+h+1)+sqrt(x+1)))$

and use the identity: $(a+b)(a-b) = a^2-b^2$

$d/dx(sqrt(x+1)) = lim_(h->0)((cancelx+h+cancel1)-(cancelx+cancel1))/(h(sqrt(x+h+1)+sqrt(x+1))$

$d/dx(sqrt(x+1)) = lim_(h->0)cancelh/(cancelh(sqrt(x+h+1)+sqrt(x+1))$

$d/dx(sqrt(x+1)) = lim_(h->0)1/(sqrt(x+h+1)+sqrt(x+1)) =1/(2sqrt(x+1))$

How do you find the derivative of sqrt( x+1 )√x+1sqrt( x+1 ) using limits?