What is the tangent of x^3+x ?

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

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tangent of x^3+x

tangent of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

{\frac{1}{( s - 4 ) ^{6}}}

\int_{1}^{7} \frac{3}{x ^{2}} d x

x \to 0 lim (x sin (\frac{2}{x}))

x \to - 2 + lim (\frac{- 1}{x + 2})

3 x^{2} - 2 cos (x)