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

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

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tangent of f(x)=(1+2x)^2,9,\at x=1

tangent of

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

y = 12 x - 3

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

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

\int \frac{ln ^{2} ( 3 x )}{x} d x

\frac{\partial}{\partial x} (s u m)

y = - 5 x - 10, (- 4, 10)

\int sin^{3} (11 x) cos^{9} (11 x) d x

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

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