What is the tangent of sqrt(x)(49.7) ?

The tangent of sqrt(x)(49.7) is y=(24.85)/(sqrt(a_{0))}x+(497sqrt(a_{0)})/(20)

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tangent of sqrt(x)(49.7)

tangent of

x (49.7)

y = \frac{24.85}{a _{0}} x + \frac{497 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 49.7)

Find the slope of

y = x 49.7 : \frac{d y}{d x} = \frac{24.85}{x}

Compute the slope of

y = x 49.7

x = a_{0} : m = \frac{24.85}{a _{0}}

Find the line with slope m=

\frac{24.85}{a _{0}}

and passing through

(a_{0}, a_{0} 49.7) : y = \frac{24.85}{a _{0}} x + \frac{497 a _{0}}{20}

y = \frac{24.85}{a _{0}} x + \frac{497 a _{0}}{20}

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

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 7, y^{^{'}} (0) = - 3

x \frac{d y}{d x} + (3 x + 1) y = e^{- 3 x}

\int \frac{1}{x ^{2} + 2 x + 65} d x

\frac{d}{d x} (3 2 x - 5)

What is the tangent of sqrt(x)(49.7) ?
The tangent of sqrt(x)(49.7) is y=(24.85)/(sqrt(a_{0))}x+(497sqrt(a_{0)})/(20)