What is the tangent of sqrt(x),\at 36 ?

The tangent of sqrt(x),\at 36 is y= 1/(2sqrt(a_{0))}x+sqrt(a_{0)}-(sqrt(a_{0)})/2

Popular >

tangent of sqrt(x),\at 36

tangent of

x, at 36

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

Compute the slope of

y = 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}

arcsec (5 ln (2 x))

\frac{d x}{d y}, y^{2} = x

\int \frac{1}{64 + ( x - 1 ) ^{2}} d x

\frac{d y}{d x} = 50 y - \frac{1}{10} y^{2}

x^{2} + 1

What is the tangent of sqrt(x),\at 36 ?
The tangent of sqrt(x),\at 36 is y= 1/(2sqrt(a_{0))}x+sqrt(a_{0)}-(sqrt(a_{0)})/2