What is the tangent of (1+2x)^2 ?

The tangent of (1+2x)^2 is y=4(1+2a_{0})x-4a_{0}^2+1

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tangent of (1+2x)^2

tangent of

(1 + 2 x)^{2}

y = 4 (1 + 2 a_{0}) x - 4 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (1 + 2 a_{0})^{2})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 4 (1 + 2 a_{0})

Find the line with slope m=

4 (1 + 2 a_{0})

and passing through

(a_{0}, (1 + 2 a_{0})^{2}) : y = 4 (1 + 2 a_{0}) x - 4 a_{0}^{2} + 1

y = 4 (1 + 2 a_{0}) x - 4 a_{0}^{2} + 1

y^{^{''}} - 8 y^{^{'}} + 41 y = 0

\frac{d}{d x} ((x^{2} - x + 1)^{- 7})

\int \frac{x - 1}{x ( x ^{2} - 5 x + 6 )} d x

\int (3 - 5 x)^{2} d x

\frac{d y}{d x} = 2 y cos (x)

What is the tangent of (1+2x)^2 ?
The tangent of (1+2x)^2 is y=4(1+2a_{0})x-4a_{0}^2+1