What is the tangent of f(x)=x^3+x,\at x=2 ?

The tangent of f(x)=x^3+x,\at x=2 is y=13x-16

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tangent of f(x)=x^3+x,\at x=2

tangent of

f (x) = x^{3} + x, at x = 2

y = 13 x - 16

Solution steps

Find the tangent point:

(2, 10)

Find the slope of

f (x) = x^{3} + x : \frac{df ( x )}{d x} = 3 x^{2} + 1

Compute the slope of

f (x) = x^{3} + x

x = 2 : m = 13

Find the line with slope m=

13

and passing through

(2, 10) : y = 13 x - 16

y = 13 x - 16

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

x \to 0 + lim (arctan (ln (0)))

\frac{\partial}{\partial x} (4 x^{2} y^{3} + 1)

\frac{d}{d x} (\frac{3 x + 2}{2 x + 3})

\int e^{2 x} sin (x) d x

What is the tangent of f(x)=x^3+x,\at x=2 ?
The tangent of f(x)=x^3+x,\at x=2 is y=13x-16