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

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

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

tangent of

f (x) = (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

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

Compute the slope of

f (x) = (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

(4 x + 5)^{3}

x \to \frac{π}{2} lim (- cot (2 x))

\frac{d}{d x} (12 tan (cosh (e^{4 x})))

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

y = e^{7 t s i n (2 t)}

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