What is the tangent of y=1sqrt(x) ?

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

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tangent of y=1sqrt(x)

tangent of

y = 1 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

y = 1 x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = 1 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}

\int_{3}^{\infty} e^{- 4 x} d x

\frac{d y}{d x} = \frac{2 y + x ^{2} - y ^{2}}{2 x}

\frac{3}{s + 1}

\int \frac{sec ^{6} ( x )}{tan ^{3} ( x )} d x

x \to 0 - lim (e^{\frac{3}{x}})

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