What is the tangent of 4x^2 ?

The tangent of 4x^2 is y=8a_{0}x-4a_{0}^2

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tangent of 4x^2

tangent of

4 x^{2}

y = 8 a_{0} x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2})

Find the slope of

y = 4 x^{2} : \frac{d y}{d x} = 8 x

Compute the slope of

y = 4 x^{2}

x = a_{0} : m = 8 a_{0}

Find the line with slope m=

8 a_{0}

and passing through

(a_{0}, 4 a_{0}^{2}) : y = 8 a_{0} x - 4 a_{0}^{2}

y = 8 a_{0} x - 4 a_{0}^{2}

n = 0 \sum \infty \frac{e ^{n}}{( n ! )}

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

\frac{\partial}{\partial x} (+ 4 x - 5 + 6 x y)

\frac{d ^{2} x}{d t ^{2}} = 5 + 2 x

\frac{d N}{d t} + N = Nt e^{t + 3}