What is the tangent of 8x-x^2 ?

The tangent of 8x-x^2 is y=(8-2a_{0})x+a_{0}^2

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tangent of 8x-x^2

tangent of

8 x - x^{2}

y = (8 - 2 a_{0}) x + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0} - a_{0}^{2})

Find the slope of

y = 8 x - x^{2} : \frac{d y}{d x} = 8 - 2 x

Compute the slope of

y = 8 x - x^{2}

x = a_{0} : m = 8 - 2 a_{0}

Find the line with slope m=

8 - 2 a_{0}

and passing through

(a_{0}, 8 a_{0} - a_{0}^{2}) : y = (8 - 2 a_{0}) x + a_{0}^{2}

y = (8 - 2 a_{0}) x + a_{0}^{2}

\frac{d}{d x} (x^{6} sin (x))

n = 0 \sum \infty \frac{1}{10 1 ^{n}}

\frac{x}{4}, 25, 0, 12

\frac{d}{d x} (x^{2} - 4 x + 3)

e^{- 4 s}