What is the tangent of 4x(3-2x)^5,\at x=2 ?

The tangent of 4x(3-2x)^5,\at x=2 is y=-84x+160

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tangent of 4x(3-2x)^5,\at x=2

tangent of

4 x (3 - 2 x)^{5}, at x = 2

y = - 84 x + 160

Solution steps

Find the tangent point:

(2, - 8)

Find the slope of

y = 4 x (3 - 2 x)^{5} : \frac{d y}{d x} = 4 ((3 - 2 x)^{5} - 10 x (3 - 2 x)^{4})

Compute the slope of

y = 4 x (3 - 2 x)^{5}

x = 2 : m = - 84

Find the line with slope m=

- 84

and passing through

(2, - 8) : y = - 84 x + 160

y = - 84 x + 160

\int \frac{x ^{2} - a ^{2}}{x ^{4}} d x

(2 x)^{^{'}}

\int \frac{( cos ^{3} ( x ) )}{3} d x

x \to 3 lim ((2 x^{2} - 4 x - 8)^{6})

\frac{\partial}{\partial x} (\frac{4 x}{y})

What is the tangent of 4x(3-2x)^5,\at x=2 ?
The tangent of 4x(3-2x)^5,\at x=2 is y=-84x+160