# differentiating trig

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#4

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Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

**emma201295**)Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

Are you talking about integration or differentiation?

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#5

**emma201295**)

Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

I hope that makes some sense!

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#6

**emma201295**)

Not unless it can be called something else too :/ I've done integration by parts, by inspection, by substitution but that's all

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(Original post by

It allows you to differentiate functions of y with respect to x by defining the derivative of y WRT x as dy/dx and then using the chain rule on these functions of y.

Are you talking about integration or differentiation?

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**majmuh24**)It allows you to differentiate functions of y with respect to x by defining the derivative of y WRT x as dy/dx and then using the chain rule on these functions of y.

Are you talking about integration or differentiation?

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(Original post by

You have sort of already done implicit differentiation; you do it whenever you normally differentiate something. Basically, wherever you would differentiate something containing a y, differentiate it as if it contains an x instead, but then multiply it by dy/dx. A normal equation where the subject is y gives a differentiated equation with dy/dx as the subject. In this case, you will then need to rearrange to find dy/dx in terms of both x's and y's.

I hope that makes some sense!

**anosmianAcrimony**)You have sort of already done implicit differentiation; you do it whenever you normally differentiate something. Basically, wherever you would differentiate something containing a y, differentiate it as if it contains an x instead, but then multiply it by dy/dx. A normal equation where the subject is y gives a differentiated equation with dy/dx as the subject. In this case, you will then need to rearrange to find dy/dx in terms of both x's and y's.

I hope that makes some sense!

I man surely the equation needs to be layed out as y=.. or x=...etc how can you dfferetiate when it is split as it is?

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#9

(Original post by

hahaha sorry differentiation the stress is clearly sending me stupid :P

**emma201295**)hahaha sorry differentiation the stress is clearly sending me stupid :P

Basically, the derivative of y WRT x is dy/dx and you treat all functions of y as a function of a function and use the chain rule on them to get a function of dy/dx, which you can then rearrange to get the derivative.

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could someone lay out steps of how to go about this as all the mathsy language is just confusing me I'm afraid

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#11

(Original post by

could someone lay out steps of how to go about this as all the mathsy language is just confusing me I'm afraid

**emma201295**)could someone lay out steps of how to go about this as all the mathsy language is just confusing me I'm afraid

Using implicit differentiation, you can get

by using the product rule on the RHS.

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(Original post by

I'll start it off.

Using implicit differentiation, you can get

by using the product rule on the RHS.

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**majmuh24**)I'll start it off.

Using implicit differentiation, you can get

by using the product rule on the RHS.

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where you've used product rule does cos2x not differentiate to -2sin2x? Sorry

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#13

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Sorry none at all, could you lay it out as steps.

I man surely the equation needs to be layed out as y=.. or x=...etc how can you dfferetiate when it is split as it is?

**emma201295**)Sorry none at all, could you lay it out as steps.

I man surely the equation needs to be layed out as y=.. or x=...etc how can you dfferetiate when it is split as it is?

So suppose I have and I need to differentiate it. Well then:

.

Take . Then we use the chain rule and get .

Since , we have .

So going back, we find the derivative is .

Of course that's not explicitly written, but you can rearrange.. You can't do any better than having y's around either. Hopefully you now see how to apply this to your problem.

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#15

(Original post by

where you've used product rule does cos2x not differentiate to -2sin2x? Sorry

**emma201295**)where you've used product rule does cos2x not differentiate to -2sin2x? Sorry

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#16

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#18

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the answer in the text book which is cos2x-2xsin2x/cosy

**emma201295**)the answer in the text book which is cos2x-2xsin2x/cosy

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#20

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Any idea how to integrate 2/x -1/(x+1)^2 -2/x+1 ?

**emma201295**)Any idea how to integrate 2/x -1/(x+1)^2 -2/x+1 ?

Latex or place brackets around the parts that you mean

Do you mean

I assume it's something to do with natural logarithms though

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