Finding the tangent equation of the curve y = x2 at x = 2______ ．

Finding the tangent equation of the curve y = x2 at x = 2______ ．

The derivative of function y = X2 is f '(x) = 2x, so the tangent slope at x = 2 is k = f' (2) = 2 × 2 = 4. When x = 2, y = 4. So the tangent equation at x = 2 is y-4 = 4 (X-2), that is, y = 4x-4. So the answer is y = 4x-4

Find the tangent equation of the square of the curve y = X,
The curve passes through the point (0, - 1),

Y = x ^ 2 = > tangent y = MX-1 simultaneous x ^ 2-mx + 1 = 0 multiple roots = > (- M) ^ 2 - 4 = 0 = > m = 2 or - 2
=> y=2x-1 or y=-2x-1.ans

Derivation of curve y = e ^ X / (e ^ x + 1) and tangent equation at x = 0

y=e^x/(e^x+1)
The tangent point is (0,1 / 2)
y‘=【e^x（e^x＋1）-e^x·e^x】/（e^x＋1）²
therefore
Slope = 1 / 4
therefore
The tangent equation is
y-1/2=1/4（x-0）
y=1/4x＋1/2