Derivatives of sec x, cosec x, cot x and tan x Watch

ScottishShortiex
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Hi,

I know the formulas/trigonometric functions for each of these but for many of the questions i cannot get the right answer!!

1) Show that the derivative of cot x is -cosec^2 x?

2) Find the derivative of a) tan 3x b) cot (tan x) ?


I'd appreciate any replies,

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Thanks
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nuodai
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Do you know the quotient rule? Because you use it to find the derivative of tan x and cot x by writing them in terms of sin x and cos x. Also, you can use the chain rule to find the derivatives of sec x and cosec x, since \sec x = (\cos x)^{-1}, and so on.
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Farhan.Hanif93
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(Original post by ScottishShortiex)
Hi,

I know the formulas/trigonometric functions for each of these but for many of the questions i cannot get the right answer!!

1) Show that the derivative of cot x is -cosec^2 x?

2) Find the derivative of a) tan 3x b) cot (tan x) ?


I'd appreciate any replies,

+ rep


Thanks
Rewrite cot(x) as \frac{\cos (x)}{\sin (x)} and use the quotient rule.
Note that Question 2 uses the chain rule.
let y=\tan (3x) and u=3x so therefore y=\tan (u) and note that \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}.
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ScottishShortiex
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(Original post by nuodai)
Do you know the quotient rule? Because you use it to find the derivative of tan x and cot x by writing them in terms of sin x and cos x. Also, you can use the chain rule to find the derivatives of sec x and cosec x, since \sec x = (\cos x)^{-1}, and so on.
Yes, i know the quotient rule, product rule and chain rule.

Your advice about the chain rule - really sure i tried that, but it got screwed up!

I'll try it again though! Thanks
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ScottishShortiex
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(Original post by Farhan.Hanif93)
Rewrite cot(x) as \frac{\cos (x)}{\sin (x)} and use the quotient rule.
Note that Question 2 uses the chain rule.
let y=\tan (3x) and u=3x so therefore y=\tan (u) and note that \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}.
I'm not sure whether when i originally tried to do this question, i over-complicated the solution or not!

i did something like tan 3x = sin 3x/cos 3x

then did like sin 3x = sin(2x+x) (which i tried to simplify and do)

likewise with cos 3x = cos (2x+x)

DO YOU NEED TO WORK THIS OUT ? ^^

thanks for your first reply
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Farhan.Hanif93
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(Original post by ScottishShortiex)
I'm not sure whether when i originally tried to do this question, i over-complicated the solution or not!

i did something like tan 3x = sin 3x/cos 3x

then did like sin 3x = sin(2x+x) (which i tried to simplify and do)

likewise with cos 3x = cos (2x+x)

DO YOU NEED TO WORK THIS OUT ? ^^

thanks for your first reply
No you don't.
Use the method I highlighted above i.e:
Let y=tan3x and u=3x.
therefore y=tanu
Find dy/du and du/dx by differentiating y=tanu and u=3x w.r.t. u and x respectively.
Then note that dy/dx=dy/du * du/dx so just multiply the two derivatives together.
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ScottishShortiex
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(Original post by Farhan.Hanif93)
No you don't.
Use the method I highlighted above i.e:
Let y=tan3x and u=3x.
therefore y=tanu
Find dy/du and du/dx by differentiating y=tanu and u=3x w.r.t. u and x respectively.
Then note that dy/dx=dy/du * du/dx so just multiply the two derivatives together.
Ok thank you so much! It's good to see where i went wrong!
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ScottishShortiex
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(Original post by Farhan.Hanif93)
Use the method I highlighted above i.e:
Let y=tan3x and u=3x.
therefore y=tanu
Find dy/du and du/dx by differentiating y=tanu and u=3x w.r.t. u and x respectively.
Then note that dy/dx=dy/du * du/dx so just multiply the two derivatives together.
I used your method and got an answer, just not the answer in the book! Can you please tell me if my answer is nearly finished, or if i've missed something OR put the wrong thing!

Here's what i put:

y = tan3x

y = sin(3x) / cos(3x) where u= 3x and y = tan u (i.e. y= sin(u)/cos(u) )

and du/dy = 3 and dy/du = cos(3x)/- sin(3x)


so therefore:

dy/dx = du/dy * dy/du

dy/dx = cos (3x)/-sin(3x) * 3

= - 3cot(3x)
THIS IS MY ANSWER!

BUT THE ANSWER IN BOOK IS: 3sec^2 (3x)

I'm so confused :confused: (only good thing is everybody in our class struggles to get the answers from the back of the book! I'm not the only one)

Any help is very appreciated!! (Also need help with question in next post where same problem occurs) i.e. different answer from book

Thanks
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Farhan.Hanif93
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(Original post by ScottishShortiex)
I used your method and got an answer, just not the answer in the book! Can you please tell me if my answer is nearly finished, or if i've missed something OR put the wrong thing!

Here's what i put:

y = tan3x

y = sin(3x) / cos(3x) where u= 3x and y = tan u (i.e. y= sin(u)/cos(u) )

and du/dy = 3 and dy/du = cos(3x)/- sin(3x)


so therefore:

dy/dx = du/dy * dy/du

dy/dx = cos (3x)/-sin(3x) * 3

= - 3cot(3x)
THIS IS MY ANSWER!

BUT THE ANSWER IN BOOK IS: 3sec^2 (3x)

I'm so confused :confused: (only good thing is everybody in our class struggles to get the answers from the back of the book! I'm not the only one)

Any help is very appreciated!! (Also need help with question in next post where same problem occurs) i.e. different answer from book

Thanks
You can leave y=tan u as it is and differentiate it from there. (consult your formula book if you don't know what the derivative of tanx is)
\frac{dy}{du}=\sec ^2u=\sec ^2(3x)
Now multiply the two together.
Basically you must have differentiate y=tanu wrongly.
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ScottishShortiex
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Next Problem:

Q. Find the derivative of.. cosec ax

Here's what i put:

y= cosec(ax)

y= 1/ sin(ax)

y = [ sin(ax) ] ^-1

dy/dx = -[sin(ax)] ^-2 * d/dx (sin ax) * d/dx (ax)

= -[sin ax] ^-2 * cos ax * a

= -acos(ax)sin(ax) ^-2
MY ANSWER!

Answer in book is: -a cosec(ax) cot (ax)

Help! :confused:
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ScottishShortiex
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(Original post by Farhan.Hanif93)
You can leave y=tan u as it is and differentiate it from there. (consult your formula book if you don't know what the derivative of tanx is)
Thanks

btw what is the derivative of tan x ? (we don't have a formula book and have never learned the derivative of tan x)
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Farhan.Hanif93
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(Original post by ScottishShortiex)
Next Problem:

Q. Find the derivative of.. cosec ax

Here's what i put:

y= cosec(ax)

y= 1/ sin(ax)

y = [ sin(ax) ] ^-1

dy/dx = -[sin(ax)] ^-2 * d/dx (sin ax) * d/dx (ax)

= -[sin ax] ^-2 * cos ax * a

= -acos(ax)sin(ax) ^-2
MY ANSWER!

Answer in book is: -a cosec(ax) cot (ax)

Help! :confused:
You're right but you haven't got it in the form the book wants:
\frac{dy}{dx} = \frac{-a\cos (ax)}{\sin ^2(ax)} = \frac{-a\cos (ax)}{\sin (ax)} \times \frac{1}{\sin (ax)} = -a\cot (ax)\csc (ax) as required.
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Farhan.Hanif93
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(Original post by ScottishShortiex)
Thanks

btw what is the derivative of tan x ? (we don't have a formula book and have never learned the derivative of tan x)
\frac{d}{dx}\tan (x) = \sec ^2(x)
You can google things like that i.e. 'what is the derivative of tanx' and you'll find out.
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ScottishShortiex
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(Original post by Farhan.Hanif93)
You're right but you haven't got it in the form the book wants:
\frac{dy}{dx} = \frac{-a\cos (ax)}{\sin ^2(ax)} = \frac{-a\cos (ax)}{\sin (ax)} \times \frac{1}{\sin (ax)} = -a\cot (ax)\csc (ax) as required.
Ok thank you, i'm just glad that i wasn't really that wrong
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ScottishShortiex
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(Original post by Farhan.Hanif93)
\frac{d}{dx}\tan (x) = \sec ^2(x)
You can google things like that i.e. 'what is the derivative of tanx' and you'll find out.
Once again, thanks for all your help! I feel like a right pain! :lol:
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Farhan.Hanif93
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(Original post by ScottishShortiex)
Once again, thanks for all your help! I feel like a right pain! :lol:
No problem. Don't feel like that, the fact that you've decided to put some effort in yourself and solve the problem first is enough to get my respect on here. Most people don't have the decency to post working. :p:
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ScottishShortiex
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(Original post by Farhan.Hanif93)
No problem. Don't feel like that, the fact that you've decided to put some effort in yourself and solve the problem first is enough to get my respect on here. Most people don't have the decency to post working. :p:
Thanks. I know :p:
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JChoudhry
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Try re-writing all these in terms of sin and cos and then use the quotient or product rule to derive them
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Congaz
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hi... can you try for me differentiating y = cosecx
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Congaz
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hi ..can you help me on finding derivative of
cosec(5x^3 )
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