substituting $A_p=A+\Delta$ into the definition $A_p^\star=\lim_{\epsilon\rightarrow 0}(A_p^\ast A_p+\epsilon I)^{-1}A_p^\ast$, and linearizing in the perturbation $\Delta$, gives $A_p^\star=A^\star+f(\Delta)$ with

$$f(\Delta)= -A^\star\Delta A^\star+ \lim_{\epsilon\rightarrow 0}(A^\ast A+\epsilon I)^{-1}\Delta^*P+{\rm order}(\Delta^2),$$

with projector $P=I-AA^\star$ onto the range of $A$. If $\Delta^\ast P=0$ the simple result

$$A_p^\star=A^\star-A^\star\Delta A^\star+{\rm order}(\Delta^2)$$

is obtained.

substituting $A_p=A+\Delta$ into the definition $A_p^\star=\lim_{\epsilon\rightarrow 0}(A_p^\ast A_p+\epsilon I)^{-1}A_p^\ast$, and linearizing in the perturbation $\Delta$, gives $A_p^\star=A^\star+f(\Delta)$ with

$$f(\Delta)= -A^\star\Delta A^\star+ \lim_{\epsilon\rightarrow 0}(A^\ast A+\epsilon I)^{-1}\Delta^*P+{\rm order}(\Delta^2),$$

with $P=I-AA^\star$ the orthogonal projector onto the range of $A$. If $\Delta^\ast P=0$ the simple result

$$A_p^\star=A^\star-A^\star\Delta A^\star+{\rm order}(\Delta^2)$$

is obtained.