What is the tangent of 8x^2 ?

The tangent of 8x^2 is y=16a_{0}x-8a_{0}^2

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tangent of 8x^2

tangent of

8 x^{2}

y = 16 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} = 16 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 16 a_{0}

Find the line with slope m=

16 a_{0}

and passing through

(a_{0}, 8 a_{0}^{2}) : y = 16 a_{0} x - 8 a_{0}^{2}

y = 16 a_{0} x - 8 a_{0}^{2}

\frac{d y}{d x} = \frac{y}{x} + e^{- \frac{y}{x}}

y = 5 x^{2} - x^{3}

x \to 2 lim (\frac{x ^{2} + 4}{x - 2})

\frac{\partial}{\partial x} (3 x^{2} \cdot e^{x^{2} y})

cos ((2 x^{2} + 1)^{7})