What is the tangent of y=(1+2x)^{11},(0,1) ?

The tangent of y=(1+2x)^{11},(0,1) is y=22x+1

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tangent of y=(1+2x)^{11},(0,1)

tangent of

y = (1 + 2 x)^{11}, (0, 1)

y = 22 x + 1

Solution steps

Find the slope of

y = (1 + 2 x)^{11} : \frac{d y}{d x} = 22 (1 + 2 x)^{10}

Compute the slope of

y = (1 + 2 x)^{11}

(0, 1) : m = 22

Find the line with slope m=

22

and passing through

(0, 1) : y = 22 x + 1

y = 22 x + 1

\frac{( 2 s - 1 )}{s ^{2} ( s + 1 ) ^{3}}

\int (x^{5} + 8 x^{3} - 9 x^{2} + 7 x - 5) d x

\int (e^{x} + \frac{1}{e ^{x}})^{2} d x

\int_{0}^{1} cos (x^{2}) d x

2 t \cdot \frac{d y}{d t} - 4 y = t

What is the tangent of y=(1+2x)^{11},(0,1) ?
The tangent of y=(1+2x)^{11},(0,1) is y=22x+1