What is the tangent of y=2x^3+x^2+23 ?

The tangent of y=2x^3+x^2+23 is y=(6a_{0}^2+2a_{0})x-4a_{0}^3-a_{0}^2+23

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tangent of y=2x^3+x^2+23

tangent of

y = 2 x^{3} + x^{2} + 23

y = (6 a_{0}^{2} + 2 a_{0}) x - 4 a_{0}^{3} - a_{0}^{2} + 23

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + a_{0}^{2} + 23)

Find the slope of

y = 2 x^{3} + x^{2} + 23 : \frac{d y}{d x} = 6 x^{2} + 2 x

Compute the slope of

y = 2 x^{3} + x^{2} + 23

x = a_{0} : m = 6 a_{0}^{2} + 2 a_{0}

Find the line with slope m=

6 a_{0}^{2} + 2 a_{0}

and passing through

(a_{0}, 2 a_{0}^{3} + a_{0}^{2} + 23) : y = (6 a_{0}^{2} + 2 a_{0}) x - 4 a_{0}^{3} - a_{0}^{2} + 23

y = (6 a_{0}^{2} + 2 a_{0}) x - 4 a_{0}^{3} - a_{0}^{2} + 23

cos^{2} (2 θ)

y^{^{'}} = - 10 y, y (0) = 5

2 y^{^{''}} + y^{^{'}} - y = x + 7

x \to - \infty lim (x \cdot 0)

\int_{- 1}^{0} t^{\frac{1}{3}} - t^{\frac{2}{3}} d t

What is the tangent of y=2x^3+x^2+23 ?
The tangent of y=2x^3+x^2+23 is y=(6a_{0}^2+2a_{0})x-4a_{0}^3-a_{0}^2+23