Revision as of 04:58, 25 July 2021

Calculus

This is a course that Henry and Ben are studying during 2021.

$(x^{n})'=n*x^{n-1}$

$[f(g(x))]'=f'(g(x))*g'(x)$

${\frac {df(g(x))}{dx}}={{df(g)} \over dg}{dg \over dx}$

@@ Line 10: / Line 10: @@
 #The sum Rule 2<math>{d (f+g) \over d x} ={d f \over d x} + {d g \over d x}</math>
-#The Difference Rule <math>(f-g)'=f'-g'</math>
+#The Difference Rule 1 <math>(f-g)'=f'-g'</math>
-#The Difference Rule <math>{d (f-g) \over d x}={d f \over d x} - {d g \over d x}</math>
+#The Difference Rule 2 <math>{d (f-g) \over d x}={d f \over d x} - {d g \over d x}</math>
 #The Product Rule <math>(f*g)'=f*g'+ g*f'</math>
+#The Product Rule <math>{d (f g) \over d x}'= f {d g \over d x} + g {d f \over d x}</math>
 #The Quotient Rule <math>({f \over g})' = {(gf'-fg') \over g^2} </math>