What is the tangent of y=(1+4x)^8,(0,1) ?

The tangent of y=(1+4x)^8,(0,1) is y=32x+1

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tangent of y=(1+4x)^8,(0,1)

tangent of

y = (1 + 4 x)^{8}, (0, 1)

y = 32 x + 1

Solution steps

Find the slope of

y = (1 + 4 x)^{8} : \frac{d y}{d x} = 32 (1 + 4 x)^{7}

Compute the slope of

y = (1 + 4 x)^{8}

(0, 1) : m = 32

Find the line with slope m=

32

and passing through

(0, 1) : y = 32 x + 1

y = 32 x + 1

x \to 0 + lim (in f (t, y, x) ini t y)

\frac{d y}{d x} + 3 y = 3 x^{2} e^{- 3 x}

\int \frac{18}{9 x ^{2} - x ^{4}} d x

x \to 0 - lim (e^{\frac{x}{x + 1}})

\int \frac{1 + x ^{2}}{x ^{2}} d x

What is the tangent of y=(1+4x)^8,(0,1) ?
The tangent of y=(1+4x)^8,(0,1) is y=32x+1