Differentiate sin 10x ?

Irish Fox

How do i diffrentiate :

sin 10x
cos 5x
tan 3x

3 tan 4x

any help would be appreciated

Improbable

http://tutorial.math.lamar.edu/Classes/CalcI/DiffTrigFcns.aspx

ChemHickey

Well, when we were in ty, we were with the class we should have been in if we hadn't done ty, and, the one thing i remember from it is how to approach tese questions!

Power first, then Trig, then finally Function.

Eg

when dealing with the first sum, note if it had a power eg, sin^2(x) or tan^5 (x) etc. Rewrite as (sinx)^2 or (tanx)^5 and then work from there (Chain rule)

Next step is to deal with the trig part, eg sin, cos, tan etc.

Finally, deal with the fnction, eg 2x or 7x etc. ( function of cos^2(3x) is 3x)

So, putting thse together, you can start, I'll use a little unrelated example containing all parts mentioned above.! :L


Differentiate Cos^2(4x) with respect to x

OK, now, remember POWER,TRIG,FUNCTION.

Rewrite as (Cos4x)^3 and now dfferentiate by chain rule.

So, "bring the 3 down to the front" ( multiply by 3) and reduce the power by 1 (3-1 = 2) so, so far you will have 3.(cos4x)^2

Next, tackle the trig. As cosx changes to -sinx, therefore, multiply your answer by -sinx. Now, you should have 3.(cos4x)^2.(-sinx)

Finally, multiply by function. Since the finction is (4x) and the dy/dx of 4x is 4, multiply your answer by the 4 to get the final answer!

So, all in all, if my calculations are correct (done in my head, it's a made up question) you should have

3.(cos4x)^2.(-sinx).(4) which easily tidies up to -12sinxcos^2(4x).

I hope it helped a biteen and my calculations are crrect!

**Timbuk2**

ChemHickey, that is not actually correct!

[latex]\frac{d}{dx}\cos^3(4x)[/latex]
[latex]=\frac{d}{dx}(\cos(4x))^3[/latex]
[latex]=3(\cos(4x))^{3-1}\frac{d}{dx}(cos(4x))[/latex]
[latex]=3(\cos(4x))^{2}(-sin(4x))\frac{d}{dx}(4x)[/latex]
[latex]=3(\cos(4x))^{2}(-sin(4x))(4)[/latex]
[latex]=-12(\cos(4x))^{2}(sin(4x))[/latex]

Expanded out slightly so you can see the chain rule!

In your solution, you have differentiated cos(x) as part of your chain rule instead of cos(4x). In words you might say

To get the derivative of [latex](\cos(4x))^3[/latex]
Derive the outer function: [latex]3(\cos(4x))^2[/latex]
Derive what is inside the brackets and multiply (note you are also doing this for this 'inner derivation'): [latex]3(\cos(4x))^2(-4sin(4x)) = -12[\cos^2(4x)][sin(4x)][/latex]

ChemHickey

Oh yes! I see that now, sorry, I just did it out of my head! :L :P Whoops, thank you for fixing it though! :L :pac: