Let the tangent equation of curve y = ax ln (x + 1) at point (0, 0) be y = 2x, then a = () A. 0B. 1C. 2D. 3

Y ′ = a − 1 x + 1, y ′ (0) = A-1 = 2, a = 3

The normal equation of y = INX in (E, 1)

y'=1/x
So tangent slope = 1 / E
So, the slope of the normal = - E
The normal equation is Y-1 = - E (x-e)
That is, ex + y-e & # 178; - 1 = 0

Let C: y = - INX (0

The derivative of y = - INX is dy = - 1 / X
The slope of L is - 1 / (e ^ - t)
The equation of L is y = - (e ^ t) x + T + 1

The tangent equation of curve y = INX at point 1,0 is?

y=x-1

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