What is the tangent of f(x)=sqrt(x),(81,9) ?

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

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tangent of f(x)=sqrt(x),(81,9)

tangent of

f (x) = x, (81, 9)

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

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x

(81, 9) : 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}

3 e^{- 5 t} + 6 e^{- 3 t}

\frac{d}{d t} (\frac{4}{t ^{2}} - \frac{5}{t} - \frac{9}{5})

f (x) = x^{10}

\int_{0}^{π} 7 sin^{2} (t co) s^{4} t d t

\frac{s + 6}{s ^{3} - 3 s ^{2} + 3 s - 1}

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