diff --git a/content/Laplace transform (lec 14-16).md b/content/Laplace transform (lec 14-16).md
index b649fc8..21c65ed 100644
--- a/content/Laplace transform (lec 14-16).md	
+++ b/content/Laplace transform (lec 14-16).md	
@@ -49,7 +49,7 @@ $\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are
 
 What if we calculate the LT of $f'$ ?
 using integration by parts:
-$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }$$=sF(s)-f(0)$
+$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f'(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+ \underbrace{s \int e^{-st}f(t) \, dt }_{ sF(s) }$$=sF(s)-f(0)$
 $\mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0)$
 in general, we can use proof by induction to show the following (try at home!):
 $\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$