What is the tangent of y=(1+3x)^9,(0,1) ?

The tangent of y=(1+3x)^9,(0,1) is y=27x+1

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tangent of y=(1+3x)^9,(0,1)

tangent of

y = (1 + 3 x)^{9}, (0, 1)

y = 27 x + 1

Solution steps

Find the slope of

y = (1 + 3 x)^{9} : \frac{d y}{d x} = 27 (1 + 3 x)^{8}

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 = 27

Find the line with slope m=

27

and passing through

(0, 1) : y = 27 x + 1

y = 27 x + 1

\frac{6}{x - 1}

\frac{d}{d x} (2 (x^{3} - x)^{4})

\frac{d}{d x} (x^{2} - x y + y^{2})

\int_{0}^{1} \frac{63}{4 y - 1} d y

e^{x} (3 x^{2} - 6 x + 6)

What is the tangent of y=(1+3x)^9,(0,1) ?
The tangent of y=(1+3x)^9,(0,1) is y=27x+1