What is the tangent of f(x)=(x^4+2x)^3,\at x=2 ?

The tangent of f(x)=(x^4+2x)^3,\at x=2 is y=40800x-73600

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tangent of f(x)=(x^4+2x)^3,\at x=2

tangent of

f (x) = (x^{4} + 2 x)^{3}, at x = 2

y = 40800 x - 73600

Solution steps

Find the tangent point:

(2, 8000)

Find the slope of

f (x) = (x^{4} + 2 x)^{3} : \frac{df ( x )}{d x} = 3 (x^{4} + 2 x)^{2} (4 x^{3} + 2)

Compute the slope of

f (x) = (x^{4} + 2 x)^{3}

x = 2 : m = 40800

Find the line with slope m=

40800

and passing through

(2, 8000) : y = 40800 x - 73600

y = 40800 x - 73600

4 e^{x + 2} + e^{- 3}

x y^{^{'}} + y = y^{2}, y (1) = - 15

y = 28 - x^{2}, y = 3 x, y = \frac{3}{5} x

\int (\frac{16}{2} x^{3} + 12 x^{5} - 2 x + 1) d x

f (x) = 5 e^{x} (cos (x) - sin (x))

What is the tangent of f(x)=(x^4+2x)^3,\at x=2 ?
The tangent of f(x)=(x^4+2x)^3,\at x=2 is y=40800x-73600