What is the tangent of y=2x(x+1)^2,(0,0) ?

The tangent of y=2x(x+1)^2,(0,0) is y=2x

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tangent of y=2x(x+1)^2,(0,0)

tangent of

y = 2 x (x + 1)^{2}, (0, 0)

y = 2 x

Solution steps

Find the slope of

y = 2 x (x + 1)^{2} : \frac{d y}{d x} = 2 (3 x^{2} + 4 x + 1)

Compute the slope of

y = 2 x (x + 1)^{2}

(0, 0) : m = 2

Find the line with slope m=

2

and passing through

(0, 0) : y = 2 x

y = 2 x

\int 3 x^{2} + 6 x + 5 d x

\frac{\partial}{\partial y} (z^{3} - x^{2} y)

(x cos (x))^{^{'}}

f (x) = x^{3} (3 - x^{2})

f (x) = \frac{x}{x ^{2} - 7}

What is the tangent of y=2x(x+1)^2,(0,0) ?
The tangent of y=2x(x+1)^2,(0,0) is y=2x