What is the tangent of f(x)=sqrt(x),\at x= 1/9 ?

The tangent of f(x)=sqrt(x),\at x= 1/9 is y= 3/2 x+1/6

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tangent of f(x)=sqrt(x),\at x= 1/9

tangent of

f (x) = x, at x = \frac{1}{9}

y = \frac{3}{2} x + \frac{1}{6}

Solution steps

Find the tangent point:

(\frac{1}{9}, \frac{1}{3})

Find the slope of

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

Compute the slope of

f (x) = x

x = \frac{1}{9} : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(\frac{1}{9}, \frac{1}{3}) : y = \frac{3}{2} x + \frac{1}{6}

y = \frac{3}{2} x + \frac{1}{6}

\frac{d}{d x} (sin (2 x))

\frac{\partial}{\partial x} (5 x^{\frac{4}{5}} y^{\frac{1}{5}})

\frac{d}{d x} (cos ((π x)^{7}))

\int \frac{6 x}{e ^{x^{2}}} d x

\frac{d}{d x} ((x^{3} - 3 x)^{4})

What is the tangent of f(x)=sqrt(x),\at x= 1/9 ?
The tangent of f(x)=sqrt(x),\at x= 1/9 is y= 3/2 x+1/6