What is the tangent of y=(1+2x)^{10},(0,1) ?

The tangent of y=(1+2x)^{10},(0,1) is y=20x+1

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tangent of y=(1+2x)^{10},(0,1)

tangent of

y = (1 + 2 x)^{10}, (0, 1)

y = 20 x + 1

Solution steps

Find the slope of

y = (1 + 2 x)^{10} : \frac{d y}{d x} = 20 (1 + 2 x)^{9}

Compute the slope of

y = (1 + 2 x)^{10}

(0, 1) : m = 20

Find the line with slope m=

20

and passing through

(0, 1) : y = 20 x + 1

y = 20 x + 1

5 x arcsin (x)

\int \frac{1 - 2 y}{y ^{2}} d y

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

\frac{d}{d x} (cos (sin (6 x)))

\frac{\partial}{\partial y} (\frac{1 + t ^{2}}{3 y - y ^{2}})

What is the tangent of y=(1+2x)^{10},(0,1) ?
The tangent of y=(1+2x)^{10},(0,1) is y=20x+1