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

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

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

tangent of

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

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

Find the slope of

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

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 = 4 (2 a_{0} + 1)

Find the line with slope m=

4 (2 a_{0} + 1)

and passing through

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

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

\frac{d ^{3}}{d x ^{3}} (cos (3 x))

\int_{1}^{\infty} \frac{1}{2 ^{x}} d x

(2 x^{2} + 1)^{3} (x^{2} - 4)^{2}

\int_{0}^{l n (3)} e^{4 x} d x

\int (4 x y) d x

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