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}

(y^{2} - 2 x y + 6 x) d x - (x^{2} - 2 x y + 2) d y = 0

\int_{1}^{e^{2}} \frac{ln ^{2} ( x ^{2} )}{x} d x

\int e^{9 x} cos (4 x) d x

\int \frac{1}{csc ( 2 x )} csc (2 x) cot (2 x) d x

\frac{d y}{d x} + 3 x^{2} y = sin (x) e^{- x^{3}}, y (0) = 1