Proof Existence and uniqueness of r(t, u)and A(t)follow from Theorem 3. 1. Hence, in order to prove differentiability and the formula for the derivative, it is sufficient to show that there exist a function C: R++R+ such that C(r)/r-0 as r-0 and E>0 such x(t,)-△(t)(u-10)-0(t)≤C(-ol enever u-pol E. Indeed, due to continuous differentiability of a, there exist C1,E0 such that a(x,t,)-a(xo(1),t,0)-4(t)(z-x0(t)-B(t)(-p0川≤C1(-0(t)+u-po|) o(p)-5(0)-△0(4-0)≤C1(l-o) whenever z-x0(t)+|a-A|≤e,t∈[to,t] Hence for ()=x(t2p)-r0(t)-△(t)(u-10) we nave 16(1)|≤C216(t)|+C3(|-po), as long as 8(t)l <1 and lu-Hol s E1, where E1>0 is sufficiently small. Together with 16(tl)≤C4(lu-0) this implies the desired bound Example 9.1 Consider the differential equation v()=1+sin(y(t),y(0)=0, where u is a small parameter. For u=0, the equation can be solved explicitly: yo(t)=t Differentiating yu(t) with respect to u at u=0 yields A(t) satisfying △()=sin(t),△(0)=0, i.e. A(t)=1-cos(t). He ence (t)=t+(1-cos(t)+O(12) for small2 Proof Existence and uniqueness of x(t, µ) and �(t) follow from Theorem 3.1. Hence, in order to prove differentiability and the formula for the derivative, it is sufficient to show that there exist a function C : R+ ∞� R+ such that C(r)/r � 0 as r � 0 and δ > 0 such that |x(t, µ) − �(t)(µ − µ0) − x0(t)| ≈ C(|µ − µ0|) whenever |µ − µ0| ≈ δ. Indeed, due to continuous differentiability of a, there exist C1, δ0 such that |a(¯x, t, µ) − a(x0(t), t, µ0) − A(t)(¯x − x0(t)) − B(t)(µ − µ0)| ≈ C1(|x¯ − x¯0(t)| + |µ − µ0|) and |x¯0(µ) − x¯0(µ0) − �0(µ − µ0)| ≈ C1(|µ − µ0|) whenever |x¯ − x¯0(t)| + |µ − µ0| ≈ δ, t ≤ [t0, t1]. Hence, for �(t) = x(t, µ) − x0(t) − �(t)(µ − µ0) we have |� ˙(t)| ≈ C2|�(t)| + C3(|µ − µ0|), as long as |�(t)| ≈ δ1 and |µ − µ0| ≈ δ1, where δ1 > 0 is sufficiently small. Together with |�(t0)| ≈ C4(|µ − µ0|), this implies the desired bound. Example 9.1 Consider the differential equation y˙(t) = 1 + µ sin(y(t)), y(0) = 0, where µ is a small parameter. For µ = 0, the equation can be solved explicitly: y0(t) = t. Differentiating yµ(t) with respect to µ at µ = 0 yields �(t) satisfying �(˙ t) = sin(t), �(0) = 0, i.e. �(t) = 1 − cos(t). Hence yµ(t) = t + µ(1 − cos(t)) + O(µ2 ) for small µ