What is the tangent of 9sqrt(x) ?

The tangent of 9sqrt(x) is y= 9/(2sqrt(a_{0))}x+9sqrt(a_{0)}-(9sqrt(a_{0)})/2

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tangent of 9sqrt(x)

tangent of

9 x

y = \frac{9}{2 a _{0}} x + 9 a_{0} - \frac{9 a _{0}}{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 a_{0})

Find the slope of

y = 9 x : \frac{d y}{d x} = \frac{9}{2 x}

Compute the slope of

y = 9 x

x = a_{0} : m = \frac{9}{2 a _{0}}

Find the line with slope m=

\frac{9}{2 a _{0}}

and passing through

(a_{0}, 9 a_{0}) : y = \frac{9}{2 a _{0}} x + 9 a_{0} - \frac{9 a _{0}}{2}

y = \frac{9}{2 a _{0}} x + 9 a_{0} - \frac{9 a _{0}}{2}

\frac{\partial ^{2}}{\partial t ^{2}} (ln (3 x + 3 c t))

e^{y} y^{^{'}} = \frac{sin ( x )}{cos ^{2} ( x )}

\frac{d}{d x} (e^{s i n (x^{2})})

y^{^{'}} - 7 y = sin (2 x)

y = 1 + x - 4 x^{\frac{1}{2}}

What is the tangent of 9sqrt(x) ?
The tangent of 9sqrt(x) is y= 9/(2sqrt(a_{0))}x+9sqrt(a_{0)}-(9sqrt(a_{0)})/2