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

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

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

tangent of

y = x, at x = 81

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

Solution steps

Find the tangent point:

(81, 9)

Find the slope of

y = x : \frac{d y}{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}

f^{^{'}} (x) = x 7 - x^{2}

x \to 3 lim (77)

\int \frac{e ^{5}}{e ^{x}} d x

\int \frac{( cos ( x ) + sin ( x ) )}{sin ( 2 x )} d x

\frac{d}{d x} (C x e^{x})

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