What is the tangent of 8-x^2 ?

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

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tangent of 8-x^2

tangent of

8 - x^{2}

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

Find the slope of

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

Compute the slope of

y = 8 - x^{2}

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

Find the line with slope m=

- 2 a_{0}

and passing through

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

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

\int 8 cos^{2} (5 x) d x

\frac{d}{d x} (\frac{5 ^{x} + 2}{5 ^{x} + 1})

y = - 8 + 5 x

\int \frac{14}{1 - cos ( 2 x )} d x

\int 6 x^{2} y - 2 x d y