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What's the derivative of sinxcosx?

 Aug 31, 2016

Best Answer 

 #1
avatar+9675 
+10

We first work out the derivative of sinx cosx

Using product rule: 

ddxsinxcosx=(cosx)(cosx)+(sinx)(sinx)=cos2x

Using chain rule:

y=u,u=sinxcosx

dydx=dydu×dudx=12u×cos2x=cos2x2sinxcosx

 Aug 31, 2016
 #1
avatar+9675 
+10
Best Answer

We first work out the derivative of sinx cosx

Using product rule: 

ddxsinxcosx=(cosx)(cosx)+(sinx)(sinx)=cos2x

Using chain rule:

y=u,u=sinxcosx

dydx=dydu×dudx=12u×cos2x=cos2x2sinxcosx

MaxWong Aug 31, 2016
 #2
avatar+118696 
+5

Thanks Max, I am just going to do it a different way.

 

What's the derivative of sinxcosx?

 

sin2x=2sinxcosxso0.5sin2x=sinxcosxddx[0.5sin2x]0.5=ddx0.5[sin2x]0.5=0.50.5[sin2x]0.52cos(2x)=0.5cos(2x)sin(2x)=cos(2x)2sin(2x)=cos(2x)22sinxcosc=cos(2x)2sinxcosc

 

There you go - our answers are the same ...   hopefully that means we are both correct  !!!

 Aug 31, 2016
 #3
avatar+9675 
0

I saw the 1 user composing an answerMelody box so I thought my answer was wrong. Thank you for saying that we are both correct XD.

MaxWong  Aug 31, 2016
 #4
avatar+44 
0

thank you both of u ;D

 Aug 31, 2016

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