What is the tangent of 2x^2,\at 2 ?

The tangent of 2x^2,\at 2 is y=4a_{0}x-2a_{0}^2

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tangent of 2x^2,\at 2

tangent of

2 x^{2}, at 2

y = 4 a_{0} x - 2 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 2 x^{2} : \frac{d y}{d x} = 4 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 = 4 a_{0}

Find the line with slope m=

4 a_{0}

and passing through

(a_{0}, 2 a_{0}^{2}) : y = 4 a_{0} x - 2 a_{0}^{2}

y = 4 a_{0} x - 2 a_{0}^{2}

\frac{d}{d x} (x e^{2 x - y^{2}})

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

\frac{\partial}{\partial x} (x^{2} y + 2 x + y)

f (x) = 6 cos (x), at x = \frac{π}{3}

\int \frac{1}{x ln ^{5} ( x )} d x

What is the tangent of 2x^2,\at 2 ?
The tangent of 2x^2,\at 2 is y=4a_{0}x-2a_{0}^2