Trigonometric derivatives

  • Thread starter Ry122
  • Start date
  • #1
565
2
How do I get the derivative of
y=tan4x
The answer in the back of the text book
says (4/cos^2(4X))
 

Answers and Replies

  • #2
mjsd
Homework Helper
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remember tan x can be written in terms of sin and cos
 
  • #3
565
2
yes i know that tanx= sinx/cosx
y=tan4x -> y=sin4x/cos4x
Then what?
 
  • #4
Gib Z
Homework Helper
3,346
5
Also use the chain rule, u=4x.
 
  • #5
565
2
y=sin4x/cos4x
u=4x
u'=4
y=sinu/cosu
y=cosu/-sinu
y=4cosu/-4sinu
y=4cos4x/-4sin4x
What am I doing wrong?
 
  • #6
Gib Z
Homework Helper
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Ahh no have you learned the quotient rule? Ask your teacher about that.

If you have a function f(x)=u/v then the derivative is f'(x)=[tex]\frac{u'v-v'u}{v^2}[/tex]. So let u=sin x and v = cos x for this, you can find the derivative of tan x is (sec x)^2, then after that use the chain rule.
 
  • #7
565
2
Using the quotient rule this is what i get.
u=sin4x
u'=4cos4x
v=cos4x
v'=-4sin4x

(-4sinX)(cos4X)--(4sinX)(sin4X)/(cos4X)^2
 
  • #8
Gib Z
Homework Helper
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No no, forget about the 4x bit for now, just use the quotient rule with u=sin x and v=cos x to find the derivative of tan x.

One you know that, THEN use the chain rule. Do you know what the chain rule is?
 
  • #10
565
2
Do you mind showing me the whole working out yourself?
That would help me alot.
Thanks
 
  • #11
The video was more than sufficient.

If you still are unable to complete the problem, seek help from your teacher, peers, or get a tutor.
 
  • #12
182
0
[tex]y=tan4x ; u=4x[/tex]
[tex]\frac{dy}{dx}=\frac{d(tanu)}{du}.\frac{du}{dx}[/tex]
[tex]\Rightarrow \ \frac{dy}{dx}= sec^2u. 4 [/tex]
[tex]\Rightarrow \ \frac{dy}{dx}=4sec^{2}4x[/tex]
 
Last edited:
  • #13
565
2
We havent learnt the other ratios yet.
The answer i need to get is in this form
(4/cos^2(4X))
I've finally figured out how to do this i just don't understand how
(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
What happens to the cos and sin squared in the numerator?
 
  • #14
hotvette
Homework Helper
996
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Hint: isn't there a trig identity relating sin2 and cos2? :smile:
 
  • #15
565
2
too easy
thanks guys
 
  • #16
182
0
We havent learnt the other ratios yet.
The answer i need to get is in this form
(4/cos^2(4X))
I've finally figured out how to do this i just don't understand how
(4cos4x^2) + (4sin4x^2) / (4cos4x^2) = (4/cos^2(4X))
What happens to the cos and sin squared in the numerator?
i thought you would know [itex] \ sec^2x=\frac{1}{cos^2x} [/itex] :uhh:
 

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