What is the tangent of y=2x^2+1 ?

The tangent of y=2x^2+1 is y=4a_{0}x-2a_{0}^2+1

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tangent of y=2x^2+1

tangent of

y = 2 x^{2} + 1

y = 4 a_{0} x - 2 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 1)

Find the slope of

y = 2 x^{2} + 1 : \frac{d y}{d x} = 4 x

Compute the slope of

y = 2 x^{2} + 1

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} + 1) : y = 4 a_{0} x - 2 a_{0}^{2} + 1

y = 4 a_{0} x - 2 a_{0}^{2} + 1

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

f (x) = sin (x) ln (x^{2} + 1)

x \to π lim (9 cot (x))

y = ln (x^{6})

What is the tangent of y=2x^2+1 ?
The tangent of y=2x^2+1 is y=4a_{0}x-2a_{0}^2+1