What is the tangent of f(x)=sqrt(x),\at x=81 ?

The tangent of f(x)=sqrt(x),\at x=81 is y= 1/18 x+9/2

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tangent of f(x)=sqrt(x),\at x=81

tangent of

f (x) = x, at x = 81

y = \frac{1}{18} x + \frac{9}{2}

Solution steps

Find the tangent point:

(81, 9)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{18}

Find the line with slope m=

\frac{1}{18}

and passing through

(81, 9) : y = \frac{1}{18} x + \frac{9}{2}

y = \frac{1}{18} x + \frac{9}{2}

\frac{ln ( 1 + x ^{2} )}{x}

\frac{\partial}{\partial x} (2 y cos (x))

\int x^{3} \cdot e^{- 2 x} d x

\int_{0}^{2.5} 80 x d x

\int 9 sin^{3} (x co) s^{5} x d x

What is the tangent of f(x)=sqrt(x),\at x=81 ?
The tangent of f(x)=sqrt(x),\at x=81 is y= 1/18 x+9/2