What is the tangent of f(x)=sqrt(x),\at x= 1/81 ?

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

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tangent of f(x)=sqrt(x),\at x= 1/81

tangent of

f (x) = x, at x = \frac{1}{81}

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

Solution steps

Find the tangent point:

(\frac{1}{81}, \frac{1}{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{9}{2}

Find the line with slope m=

\frac{9}{2}

and passing through

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

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

5 \frac{1}{x}

\int_{0}^{3.5} 90 x d x

n = 1 \sum \infty \frac{1}{2 n ^{2}}

x^{2} e^{x} - x^{3} - x^{2}

\frac{d}{d t} (2 e^{3 t})

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