What is the tangent of 9x-8sqrt(x) ?

The tangent of 9x-8sqrt(x) is y=(9-4/(sqrt(a_{0))})x-4sqrt(a_{0)}

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tangent of 9x-8sqrt(x)

tangent of

9 x - 8 x

y = (9 - \frac{4}{a _{0}}) x - 4 a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 a_{0} - 8 a_{0})

Find the slope of

y = 9 x - 8 x : \frac{d y}{d x} = 9 - \frac{4}{x}

Compute the slope of

y = 9 x - 8 x

x = a_{0} : m = 9 - \frac{4}{a _{0}}

Find the line with slope m=

9 - \frac{4}{a _{0}}

and passing through

(a_{0}, 9 a_{0} - 8 a_{0}) : y = (9 - \frac{4}{a _{0}}) x - 4 a_{0}

y = (9 - \frac{4}{a _{0}}) x - 4 a_{0}

(\frac{ln ^{2} ( x )}{x})^{^{'}}

x \to 2 lim (cos (x - 2))

x \to \infty lim (1 + \frac{x}{2})

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

\frac{\partial}{\partial y} (e^{2 x + y})

What is the tangent of 9x-8sqrt(x) ?
The tangent of 9x-8sqrt(x) is y=(9-4/(sqrt(a_{0))})x-4sqrt(a_{0)}