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

The tangent of y=x^3+x is y=(3a_{0}^2+1)x-2a_{0}^3

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tangent of y=x^3+x

tangent of

y = x^{3} + x

y = (3 a_{0}^{2} + 1) x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x^{3} + x : \frac{d y}{d x} = 3 x^{2} + 1

Compute the slope of

y = x^{3} + x

x = a_{0} : m = 3 a_{0}^{2} + 1

Find the line with slope m=

3 a_{0}^{2} + 1

and passing through

(a_{0}, a_{0}^{3} + a_{0}) : y = (3 a_{0}^{2} + 1) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} + 1) x - 2 a_{0}^{3}

\int x^{2} e x d x

\frac{sin ( θ )}{3} + \frac{c}{θ}

3 z

\frac{d}{d x} (\frac{1}{4} x^{4})

\int 3 x cos (4 x) d x

What is the tangent of y=x^3+x ?
The tangent of y=x^3+x is y=(3a_{0}^2+1)x-2a_{0}^3