What is the tangent of sqrt(x)(9.3) ?

The tangent of sqrt(x)(9.3) is y=(4.65)/(sqrt(a_{0))}x+(93sqrt(a_{0)})/(20)

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tangent of sqrt(x)(9.3)

tangent of

x (9.3)

y = \frac{4.65}{a _{0}} x + \frac{93 a _{0}}{20}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} \cdot 9.3)

Find the slope of

y = x 9.3 : \frac{d y}{d x} = \frac{4.65}{x}

Compute the slope of

y = x 9.3

x = a_{0} : m = \frac{4.65}{a _{0}}

Find the line with slope m=

\frac{4.65}{a _{0}}

and passing through

(a_{0}, a_{0} 9.3) : y = \frac{4.65}{a _{0}} x + \frac{93 a _{0}}{20}

y = \frac{4.65}{a _{0}} x + \frac{93 a _{0}}{20}

y^{^{'}} = - \frac{y}{t} + 2

f (x) = x^{4} - 4 x^{2}, g (x) = x^{2} - 4

\int (6 x + 2) d x

x \to - \infty lim (arccot (x))

\frac{d}{d x} (e - x)

What is the tangent of sqrt(x)(9.3) ?
The tangent of sqrt(x)(9.3) is y=(4.65)/(sqrt(a_{0))}x+(93sqrt(a_{0)})/(20)