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

The tangent of y=(1+2x)^2,\at x=1 is y=12x-3

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

tangent of

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

y = 12 x - 3

Solution steps

Find the tangent point:

(1, 9)

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 = 1 : m = 12

Find the line with slope m=

12

and passing through

(1, 9) : y = 12 x - 3

y = 12 x - 3

\int \frac{1}{- 3 x ^{2} + 3 x + 7} d x

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

\int \frac{1}{5 - 4 x - x ^{2}} d x

\int cos (4 x) d x

\int \frac{( x + 1 )}{x ^{2} - 4 x + 12} d x

What is the tangent of y=(1+2x)^2,\at x=1 ?
The tangent of y=(1+2x)^2,\at x=1 is y=12x-3