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

The tangent of f(x)=sqrt(x) is y= 1/(2sqrt(a_{0))}x+sqrt(a_{0)}-(sqrt(a_{0)})/2

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tangent of f(x)=sqrt(x)

tangent of

f (x) = x

y = \frac{1}{2 a _{0}} x + a_{0} - \frac{a _{0}}{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0})

Find the slope of

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

Compute the slope of

f (x) = x

x = a_{0} : m = \frac{1}{2 a _{0}}

Find the line with slope m=

\frac{1}{2 a _{0}}

and passing through

(a_{0}, a_{0}) : y = \frac{1}{2 a _{0}} x + a_{0} - \frac{a _{0}}{2}

y = \frac{1}{2 a _{0}} x + a_{0} - \frac{a _{0}}{2}

\frac{\partial}{\partial x} (\frac{4 x + 3 t}{x - 2 t})

2 x^{4} ln (x)

\int_{0}^{1} π (x^{2} - x) d x

\int_{0}^{1} \frac{( 1 + e ^{x} )}{e ^{x}} d x

\int \frac{x ^{2}}{9 - 4 x ^{2}} d x

What is the tangent of f(x)=sqrt(x) ?
The tangent of f(x)=sqrt(x) is y= 1/(2sqrt(a_{0))}x+sqrt(a_{0)}-(sqrt(a_{0)})/2