What is the tangent of 2x(5-4x)^4,\at x=1 ?

The tangent of 2x(5-4x)^4,\at x=1 is y=-30x+32

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tangent of 2x(5-4x)^4,\at x=1

tangent of

2 x (5 - 4 x)^{4}, at x = 1

y = - 30 x + 32

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

y = 2 x (5 - 4 x)^{4} : \frac{d y}{d x} = 2 ((5 - 4 x)^{4} - 16 x (5 - 4 x)^{3})

Compute the slope of

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

x = 1 : m = - 30

Find the line with slope m=

- 30

and passing through

(1, 2) : y = - 30 x + 32

y = - 30 x + 32

(- 3 x y^{2} + 5 y) + (- 3 x^{2} y + 5 x) y^{^{'}} = 0

\frac{d}{d x} ((x^{2} - 4)^{\frac{1}{2}})

\int p^{2} d p

x \to - 2 lim (\frac{1}{x + 2})

e^{4}

What is the tangent of 2x(5-4x)^4,\at x=1 ?
The tangent of 2x(5-4x)^4,\at x=1 is y=-30x+32