Revision as of 05:05, 25 July 2021

Calculus

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

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

The Quotient Rule 2 ${d({f \over g}) \over dx}={g{df \over dx}-f{dg \over dx} \over g^{2}}$

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

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

@@ Line 13: / Line 13: @@
 #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 1<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 Product Rule 2<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>
+#The Quotient Rule 1<math>({f \over g})' = {(gf'-fg') \over g^2} </math>
-#The Quotient Rule <math>{d ({f \over g})  \over d x} = {  g {d f \over d x} - f {d g \over d x} \over g^2} </math>
+#The Quotient Rule 2<math>{d ({f \over g})  \over d x} = {  g {d f \over d x} - f {d g \over d x} \over g^2} </math>
 ====[[Calculus:Derivative of Trigonometric Functions|Derivative of Trigonometric Functions]]====