If f (x) = SiNx / (1 + SiNx), then the tangent slope of F (x) at x = 0 is

If f (x) = SiNx / (1 + SiNx), then the tangent slope of F (x) at x = 0 is

Solution derivation f '(x) = [SiNx / (1 + SiNx)]'
=[（sinx）′（1+sinx)-（sinx）（1+sinx)′]/(1+sinx)²
=[cosx（1+sinx)-（sinx）cosx]/(1+sinx)²
When x = 0, f ′ (0) = [cos0 (1 + sin0) - (sin0) cos0] / (1 + sin0) & # 178; = 1 / 1 = 1
That is, (x) = SiNx / (1 + SiNx), then the tangent slope of F (x) at x = 0 is 1

Let f (x) = SiNx − 3cosx + X + 1. (I) find the tangent equation of F (x) at x = 0; (II) note that the opposite side lengths of a, B and C of △ ABC are a, B and C respectively, f '(b) = 3 and a + C = 2, and find the minimum of side length B

(I) when x = 0, f (0) = 1-3, then the tangent point is (0, 1-3) ∵ f ′ (x) = cosx + 3sinx + 1, ∥ f ′ (0) = 2 ∥ the tangent equation of function f (x) at x = 0 is Y - (1-3) = 2 (x-0), i.e. y = 2x + (1-3); (II) from (I) f ′ (b) = 2Sin (B + π 6) + 1 = 3, i.e. sin (B + π 6) = 1, ∥ B = π 3, B2 = A2 + c2-2accosb = A2 + c2-ac = (a + C) 2-3ac = 4-3 AC ≥ 4-3 · (a + C2) 2 = 4-3 = 1, if and only if a = C = 1, take the equal sign ∵ B2 ≥ 1, ∵ b > 0, ∵ B ≥ 1, ∵ bmin = 1

Find the average rate of change (x ≠ 0) of function y = 1 / X between x0 and x0 + △ x

F (x0 ＋△ x) - f (x0) = 1 / (x0 ＋△ x) - 1 / (x0) = (- △ x) / [(x0) (x0 ＋△ x)], so the average change rate is:
[f(x0＋△x)－f(x0)]/(△x)=－1/[(x0)(x0＋△x)].

Find the average change rate of function y = f (x) = 1 / radical X in the interval [1,1 + △ x]

Average change = △ Y / △ x = (1 / √ (x + △ x) - 1 / √ x) / △ x, the number of X and △ x can be replaced as much as possible
When △ X - > 0, it is the concept of derivative
y’=△y / △x=(1/√(x+△x)- 1/√x)/ △x=[√(x+△x)-(1+△x)] / [△x (x+△x)]
When √ (x + △ x) = 1 + △ X / 2 (when △ X - > 0)
Y '= [(1 + △ X / 2) - (1 + △ x)] / [△ x (1 + △ x)] = - △ X / 2 / [△ x (1 + △ x)]
=-1/2 / （1+△x）=-1/2
Y=1/√x=x^(-1/2)
Y '= - 1 / 2 x ^ (- 1 / 2-1), when x = 1, y' = - 1 / 2