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

The tangent of (x^3+2x)^3,\at x=4 is y=777600x-2737152

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

tangent of

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

y = 777600 x - 2737152

Solution steps

Find the tangent point:

(4, 373248)

Find the slope of

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

Compute the slope of

y = (x^{3} + 2 x)^{3}

x = 4 : m = 777600

Find the line with slope m=

777600

and passing through

(4, 373248) : y = 777600 x - 2737152

y = 777600 x - 2737152

\int_{0}^{2} x^{6} d x

\int x cos (4 π) x^{2} d x

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

y = 2 x^{2}, y = 2 x + 6

ln (x)

What is the tangent of (x^3+2x)^3,\at x=4 ?
The tangent of (x^3+2x)^3,\at x=4 is y=777600x-2737152