What is the tangent of f(x)=sqrt(x),\at x=36 ?

The tangent of f(x)=sqrt(x),\at x=36 is y= 1/12 x+3

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tangent of f(x)=sqrt(x),\at x=36

tangent of

f (x) = x, at x = 36

y = \frac{1}{12} x + 3

Solution steps

Find the tangent point:

(36, 6)

Find the slope of

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

Compute the slope of

f (x) = x

x = 36 : m = \frac{1}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(36, 6) : y = \frac{1}{12} x + 3

y = \frac{1}{12} x + 3

f (x) = 4 x^{3} + 9

(- x^{- 2})^{^{'}}

(x^{2} + 2 x - 1) y^{^{'}} - (x + 1) y = x - 1

x \to 0 lim (x^{4} cos (\frac{5}{x}))

\int t^{2} (t^{3} - 4)^{2} d t

What is the tangent of f(x)=sqrt(x),\at x=36 ?
The tangent of f(x)=sqrt(x),\at x=36 is y= 1/12 x+3