What is the tangent of 9x^2 ?

The tangent of 9x^2 is y=18a_{0}x-9a_{0}^2

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tangent of 9x^2

tangent of

9 x^{2}

y = 18 a_{0} x - 9 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 9 x^{2} : \frac{d y}{d x} = 18 x

Compute the slope of

y = 9 x^{2}

x = a_{0} : m = 18 a_{0}

Find the line with slope m=

18 a_{0}

and passing through

(a_{0}, 9 a_{0}^{2}) : y = 18 a_{0} x - 9 a_{0}^{2}

y = 18 a_{0} x - 9 a_{0}^{2}

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

\int \frac{ln ( 2 )}{x ^{2}} d x

y^{^{'}} = \frac{( 3 t ^{2} - e ^{t} )}{( 2 y - 5 )}, y (0) = 1

n = 1 \sum \infty ln (n + 2)

x \to 2 - lim (ln (x^{2} (3 - x)))