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}

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 = 32

Find the line with slope m=

32

and passing through

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

y = 32 x + 1

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