Difference between revisions of "Calculus:Derivative of Trigonometric Functions"

Revision as of 12:32, 31 August 2021

$f(sinx)'=cos(x)$ and $f(cosx)'=-sin(x)$ you only can tell by looking at the graph so we will skip it to.

So we will started with $f(tanx)'=sec^{2}(x)$

We know that $tanx={sinx \over cosx}$ If you have learn trigonometry then.

$f({sinx \over cosx})'$ by using the Quotient Rule $({f \over g})'={(gf'-fg') \over g^{2}}$

$f({sinx \over cosx})'={{cosx(sinx)'-sinx(cosx)'} \over {cosx}^{2}}$

$f({sinx \over cosx})'={{cosxcosx-sinx(-sinx)} \over {cosx}^{2}}$

$f({sinx \over cosx})'={{{cos}^{2}x+{sin}^{2}x} \over {cosx}^{2}}$

Failed to parse (syntax error): {\displaystyle {cos}^2 x + {sin}^2 x} = 1}

$f({sinx \over cosx})'={1 \over {cosx}^{2}}$

@@ Line 20: / Line 20: @@
 <math>f({sin x \over cos x})'={{cos x (sin x)' - sin x (cos x)'} \over {cos x}^2 }</math>
+<math>f({sin x \over cos x})'={{cos x cos x - sin x (-sin x)} \over {cos x}^2 }</math>
+<math>f({sin x \over cos x})'={{{cos}^2 x + {sin}^2 x} \over {cos x}^2 }</math>
+<math>{cos}^2 x + {sin}^2 x} = 1</math>
+<math>f({sin x \over cos x})'={1 \over {cos x}^2 }</math>