How do you find the slope of the tangent line to the graph of the equation at the point with x-coordinate for $y = \sqrt{x}$ $y = sqrt(x)$ at p(4,2)?

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Answer 1 · 2015-06-10T14:02:19

You first find the derivative and then use that at $x_{p} = 4$ $x_p=4$

$y = \sqrt{x} = x^{\frac{1}{2}} \to$ $y=sqrtx=x^(1/2)->$

$y'=1/2 x^(-1/2)=1/(2*sqrtx)$

Using $x_p=4$ :

$y'(4)=1/(2.sqrt4)=1/4$

And this is the slope.

How do you find the slope of the tangent line to the graph of the equation at the point with x-coordinate for y = sqrt(x)y=√xy = sqrt(x) at p(4,2)?