What is the Derivative of y = √x at x = 1?

[fiziksfun]

I was asked to find the derivative of y=sqrtx at x=1

I need to use the formula

lim x -> a = f(x) - f(a) / (x-a) to find the derivative at x=1

I have..

sqrtx - 1 / (x - 1) BUT i can't simply it from there. Help!

THANKSZ.

 

[arildno]

Okay:
$$\frac{\sqrt{x}-1}{x-1}=\frac{\sqrt{x}+1}{\sqrt{x}+1}\frac{\sqrt{x}-1}{x-1}=\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{x-1}\frac{1}{\sqrt{x}+1}$$
Does that help you a bit?

 

[tacman]

To find the derivative of y = √x at x = 1, we can use the limit definition of derivative as you have mentioned. Let's plug in x = 1 into the formula:

lim x->1 = √x - √1 / (x - 1)

Since √1 = 1, we can simplify the numerator to:

lim x->1 = √x - 1 / (x - 1)

Now, to simplify further, we can use the conjugate of the numerator (which is √x + 1) to eliminate the radical in the numerator:

lim x->1 = (√x - 1)(√x + 1) / (x - 1)(√x + 1)

= (x - 1) / (x - 1)(√x + 1)

= 1 / (√x + 1)

Finally, we can plug in x = 1 to find the derivative at x = 1:

lim x->1 = 1 / (√1 + 1)

= 1/2

Therefore, the derivative of y = √x at x = 1 is 1/2. I hope this helps!