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

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

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tangent of 8x-8sqrt(x)

tangent of

8 x - 8 x

y = (8 - \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}, 8 a_{0} - 8 a_{0})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8 - \frac{4}{a _{0}}

Find the line with slope m=

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

and passing through

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

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

\int_{0.5}^{3} \frac{sin ( x )}{x} d x

t - 3

y = - 4 cos (x), at x = \frac{π}{2}

\int (x + 6) (x - 3) d x

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

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