What is the tangent of 2x(x^2-4x+5)^8,\at x=2 ?

The tangent of 2x(x^2-4x+5)^8,\at x=2 is y=2x

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tangent of 2x(x^2-4x+5)^8,\at x=2

tangent of

2 x (x^{2} - 4 x + 5)^{8}, at x = 2

y = 2 x

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

y = 2 x (x^{2} - 4 x + 5)^{8} : \frac{d y}{d x} = 2 ((x^{2} - 4 x + 5)^{8} + 8 x (x^{2} - 4 x + 5)^{7} (2 x - 4))

Compute the slope of

y = 2 x (x^{2} - 4 x + 5)^{8}

x = 2 : m = 2

Find the line with slope m=

2

and passing through

(2, 4) : y = 2 x

y = 2 x

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 t^{3}

(\frac{2}{x ^{3}})^{^{'}}

y = \frac{1}{3} x^{3}

\int e^{- 0.2 x} d x

\int \frac{x}{( x ^{2} + 1 ) ^{\frac{15}{2}}} d x

What is the tangent of 2x(x^2-4x+5)^8,\at x=2 ?
The tangent of 2x(x^2-4x+5)^8,\at x=2 is y=2x