What is the tangent of f(x)=(x^3+6x)^4,\at x=2 ?

The tangent of f(x)=(x^3+6x)^4,\at x=2 is y=576000x-992000

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tangent of f(x)=(x^3+6x)^4,\at x=2

tangent of

f (x) = (x^{3} + 6 x)^{4}, at x = 2

y = 576000 x - 992000

Solution steps

Find the tangent point:

(2, 160000)

Find the slope of

f (x) = (x^{3} + 6 x)^{4} : \frac{df ( x )}{d x} = 4 (x^{3} + 6 x)^{3} (3 x^{2} + 6)

Compute the slope of

f (x) = (x^{3} + 6 x)^{4}

x = 2 : m = 576000

Find the line with slope m=

576000

and passing through

(2, 160000) : y = 576000 x - 992000

y = 576000 x - 992000

\int_{0}^{2} \frac{2 x}{( x ^{2} + 3 ) ^{3}} d x

x^{2} + x

sin (\frac{π x}{3})

\int \frac{( 4 + x + x ^{2} )}{x} d x

f (x) = ln (\frac{x ^{2} - 1}{x ^{2} + 1})

What is the tangent of f(x)=(x^3+6x)^4,\at x=2 ?
The tangent of f(x)=(x^3+6x)^4,\at x=2 is y=576000x-992000