What is the tangent of y=2x^2 ?

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

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tangent of y=2x^2

tangent of

y = 2 x^{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

Compute the slope of

y = 2 x^{2}

x = a_{0} : 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}

\int_{0}^{y + z} \frac{1}{( x + y + z ) ^{3}} d x

e^{x} (x - 1)

y^{^{'}} + sin (t) y = sin (t), y (0) = 6

\frac{\partial}{\partial x} ((x^{2} - 2 x) cos (y))

\frac{d}{d x} ((e^{3 x}) (cos (2 x)))