Chris: Difference between revisions

Revision as of 17:13, 28 March 2023

Je m'appelle Christopher Sanchez.
$f(x)=2x-3$
$g(x)=x+2$
$(f(g)),(f-g),(f+g),({\frac {f}{x}})$

${\frac {d}{dx}}[c]=0$
${\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]$

${\frac {d}{dx}}[f(x)\pm g(x)]={\frac {d}{dx}}[f(x)]\pm {\frac {d}{dx}}[g(x)]$

${\frac {d}{dx}}[{\frac {f}{g}}]=F'(x)\cdot g(x)-g'(x)\cdot {\frac {f}{g^{2}}}$
${\frac {d}{dx}}[c]=0$

@@ Line 3: / Line 3: @@
 <math>g(x)= x+2 </math><br>
 <math> (f(g)),(f-g),(f+g),({\frac{f}{x}}) </math>
+===Review===
 <math>{\frac{d}{dx}} [c] = 0 </math> <br>
 <math>{\frac{d}{dx}} [c\cdot f(x)] = c\cdot{\frac{d}{dx}} [f(x)] </math> <br>
 <math>{\frac{d}{dx}} [f(x)\pm g(x)] = {\frac{d}{dx}} [f(x)] \pm {\frac{d}{dx}} [g(x)] </math> <br>
-<math>{\frac{d}{dx}} [c] = 0 </math> <br>
+===Quotient Rule===
+<math>{\frac{d}{dx}} [{\frac{f}{g}}] = F'(x) \cdot g(x) - g'(x) \cdot {\frac{f}{g^2}} </math> <br>
 <math>{\frac{d}{dx}} [c] = 0 </math> <br>