What is the tangent of f(x)=sqrt(x),\at x=2 ?

The tangent of f(x)=sqrt(x),\at x=2 is y= 1/(2sqrt(2))x+sqrt(2)-1/(sqrt(2))

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tangent of f(x)=sqrt(x),\at x=2

tangent of

f (x) = x, at x = 2

y = \frac{1}{2 2} x + 2 - \frac{1}{2}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = x : \frac{df ( x )}{d x} = \frac{1}{2 x}

Compute the slope of

f (x) = x

x = 2 : m = \frac{1}{2 2}

Find the line with slope m=

\frac{1}{2 2}

and passing through

(2, 2) : y = \frac{1}{2 2} x + 2 - \frac{1}{2}

y = \frac{1}{2 2} x + 2 - \frac{1}{2}

\frac{d y}{d t} + e^{t} y = - 3 e^{t}

sin (t) cos (t)

\frac{\partial}{\partial x} (5 x y e^{(x + y)})

\frac{d}{d x} (2 x + \frac{4 x}{3 x - 1})

f (x) = \frac{4 x}{x ^{2} - 5}, at x = 3

What is the tangent of f(x)=sqrt(x),\at x=2 ?
The tangent of f(x)=sqrt(x),\at x=2 is y= 1/(2sqrt(2))x+sqrt(2)-1/(sqrt(2))