What is the tangent of x^2+9 ?

The tangent of x^2+9 is y=2a_{0}x-a_{0}^2+9

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tangent of x^2+9

tangent of

x^{2} + 9

y = 2 a_{0} x - a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 9)

Find the slope of

y = x^{2} + 9 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 9

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 9) : y = 2 a_{0} x - a_{0}^{2} + 9

y = 2 a_{0} x - a_{0}^{2} + 9

\int_{1}^{5} \frac{x ^{2} + 8}{6 x - x ^{2}} d x

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

\int \frac{cot ^{2} ( x )}{cos ^{2} ( x )} d x

\frac{d}{d x} (\frac{1}{4} (e^{2 x} + e^{- 2 x}))

\frac{d}{d x} (\frac{x ^{3} - x ^{2} + 1}{5})