What is the tangent of y=(1+2x)^2,(1,9) ?

The tangent of y=(1+2x)^2,(1,9) is y=12x-3

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tangent of y=(1+2x)^2,(1,9)

tangent of

y = (1 + 2 x)^{2}, (1, 9)

y = 12 x - 3

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 12

Find the line with slope m=

12

and passing through

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

y = 12 x - 3

\frac{\partial}{\partial x} (arctan (y x))

\frac{d}{d x} (1 + x)

x y^{^{'}} - y = x^{2}

x \to 7 lim (x - 7)

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

What is the tangent of y=(1+2x)^2,(1,9) ?
The tangent of y=(1+2x)^2,(1,9) is y=12x-3