What is the tangent of x^2+8 ?

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

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tangent of x^2+8

tangent of

x^{2} + 8

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x^{2} + 8

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

Find the line with slope m=

2 a_{0}

and passing through

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

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

n = 1 \sum \infty \frac{1}{4 n ^{6} + n + 1}

\frac{d y}{d x} = \frac{9 x}{y}

\int (\frac{1}{x \cdot ln ( x )}) d x

\int \frac{1}{x ^{2} - 4 x + 5} d x

x^{2} y^{^{'}} + 2 x y = 15 cos^{2} (x)