More differentiation problems

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Main Question or Discussion Point

1. If [itex]x = \cos \theta[/itex] and [itex]y = \cos \theta + \sin \theta[/itex], [itex]\frac{dy}{dx}[/itex] is? (Answer: [itex]1 - \cot \theta[/itex])

I'm just not seeing this question... any help appreciated.

2. [itex]y = \ln(\sec 2x + \tan 2x)[/itex] (Answer = [itex]2\sec 2x[/itex])

I get to [tex]\frac{1}{(2\sec x\tan x) + (2\sec^2 x)}[/tex] but it doesn't seem that I'm on the right track compared to the answers I have.
 
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Answers and Replies

TD
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1) If [itex]x = \cos \theta[/itex] then [itex]dx = - \sin \theta[/itex], right?
Now compute [itex]dy[/itex] as well and do the division :wink:

2) What do you have to do here? Just compute [itex]dy[/itex] or [itex]dy/dx[/itex] again with the x from 1?
 
783
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TD said:
1) If [itex]x = \cos \theta[/itex] then [itex]dx = - \sin \theta[/itex], right?
Now compute [itex]dy[/itex] as well and do the division :wink:

2) What do you have to do here? Just compute [itex]dy[/itex] or [itex]dy/dx[/itex] again with the x from 1?
Thanks I got 1 now! As for 2, it's a seperate question. I just need to differentiate with respect to x.
 
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For your second one, you need to remember that [tex]\frac{d(\ln{x})}{dx}=\frac{1}{x}*dx[/tex] Just remember to use the chain rule and you should get the right answer.
 
TD
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For the derivative of an ln of a function, you can use the following (follows from the chain rule)

[tex]\left(\ln \left( {f\left( x \right)} \right)\right)^\prime = \frac{{f'\left( x \right)}}
{{f\left( x \right)}}[/tex]
 
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OK, so I should get [tex]\frac{(2 \sec 2x \tan 2x) + (2 \sec^2 2x)}{(\sec 2x + \tan 2x)}[/tex] but I still can't get it to [itex]2 \sec 2x[/tex], my algebra is failing me! :blushing:
 
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TD
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For the derivative

[tex]\left( {\sec \left( {2x} \right) + \tan \left( {2x} \right)} \right)^\prime = \frac{{2\sin \left( {2x} \right)}}{{\cos ^2 \left( {2x} \right)}} + \frac{2}{{\cos ^2 \left( {2x} \right)}} = \frac{{2\sin \left( {2x} \right) + 2}}{{\cos ^2 \left( {2x} \right)}}[/tex]

Then it's just putting everything into sines and cosines

[tex]\frac{{\frac{{2\sin \left( {2x} \right) + 2}}{{\cos ^2 \left( {2x} \right)}}}}{{\sec \left( {2x} \right) + \tan \left( {2x} \right)}} = \frac{{\frac{{2\sin \left( {2x} \right) + 2}}{{\cos ^2 \left( {2x} \right)}}}}{{\frac{1}{{\cos \left( {2x} \right)}} + \frac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}}} = \frac{{\frac{{2\left( {1 + \sin \left( {2x} \right)} \right)}}{{\cos ^2 \left( {2x} \right)}}}}{{\frac{{1 + \sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}}} = \frac{2}{{\cos \left( {2x} \right)}} = 2\sec \left( {2x} \right)[/tex]
 
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Wow, I never would have figured it out thinking the way I was. Thank you for being patient with me :smile:
 
TD
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No problem.

Normally I'd have tried to guide you more through it but I hope my previous posts is clear enough and that you actually understand what happened.

Since you try to simpplify, it is easier to lose the tangent and secans and switch all over to sines and cosines. Then it's purely algebra, simplifying the expression.
 
Fermat
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cscott said:
OK, so I should get [tex]\frac{(2 \sec 2x \tan 2x) + (2 \sec^2 2x)}{(\sec 2x + \tan 2x)}[/tex] but I still can't get it to [itex]2 \sec 2x[/tex], my algebra is failing me! :blushing:
take out [itex]2 \sec 2x[/tex] as a factor from the numerator. The other bits will cancel.
 
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Fermat said:
take out [itex]2 \sec 2x[/tex] as a factor from the numerator. The other bits will cancel.
Ah, I never thought of that either. Thanks.
 

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