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# Help with differentiating... Watch

1. x = cox y/2 + 1

Find dx/dy when y = 2

I differentiated it and got -sin y/2 but apparently it's - 1/2 sin y/2 ... I don't get where the -1/2 comes from as cos differentiated is -sin and the 1 disappears...

Probably missing something simple here but I can't see what. Thanks!
2. When differentiating a trigonometric function you need to differentiate what's within the function and multiply the result by it. In this case d/dy(y/2)= 1/2. This is where the half comes from.
3. You need to use the chain rule here. Let .
4. (Original post by nuodai)
You need to use the chain rule here. Let .
Ah, I see that, thank you for your help!

If I could just ask something else, do you know how I would go about differentiating x ln (4x - 3) ?

5. (Original post by themagicpiano)
Ah, I see that, thank you for your help!

If I could just ask something else, do you know how I would go about differentiating x ln (4x - 3) ?

You can use product rule where u=x and v=ln(4x-3), you know how to differentiate ln functions right?
6. (Original post by themagicpiano)
Ah, I see that, thank you for your help!

If I could just ask something else, do you know how I would go about differentiating x ln (4x - 3) ?

product rule!

let u = x and v = ln(4x-3)

then dy/dx = u(dv/dx)+v(du/dx)
7. well this is how it goes

x=cos y/2 is the same as:

x= cos 1/2y

so to differentiate
cos differentiates to -sin

the coefficient of y (1/2) is bought to the front

so you get dx/dy = -1/2sin y/2

Hope this helps
8. (Original post by roar558)
You can use product rule where u=x and v=ln(4x-3), you know how to differentiate ln functions right?
Yeah, thank you and thank you @the greatest. I just got a bit confused about how to split them up because of the ln. It just confused me. Thanks for your help.
9. (Original post by viksta1000)
well this is how it goes

x=cos y/2 is the same as:

x= cos 1/2y

so to differentiate
cos differentiates to -sin

the coefficient of y (1/2) is bought to the front

so you get dx/dy = -1/2sin y/2

Hope this helps
Great explanation, thank you! Makes it so much clearer!

So the coefficient always gets moved to the front when differentiating?

Thanks
10. (Original post by themagicpiano)
Great explanation, thank you! Makes it so much clearer!

So the coefficient always gets moved to the front when differentiating?

Thanks
only in trig functions...you can use that as a general rule yes
11. (Original post by themagicpiano)
Great explanation, thank you! Makes it so much clearer!

So the coefficient always gets moved to the front when differentiating?

Thanks
(Original post by viksta1000)
only in trig functions...you can use that as a general rule yes
As it happens it's fine for any function.

Suppose instead of you have . Then if we set we have , so and . Putting this together with the chain rule, we get . So, if you have a function of something in the form , you can just take the derivative as you would normally and then multiply by .
12. (Original post by nuodai)
As it happens it's fine for any function.

Suppose instead of you have . Then if we set we have , so and . Putting this together with the chain rule, we get . So, if you have a function of something in the form , you can just take the derivative as you would normally and then multiply by .
i stand corrected

i just didn't want the OP to get confused when it comes to logs

e.g. ln2x => 1/x or 3ln2x => 3/x etc
13. (Original post by viksta1000)
i stand corrected

i just didn't want the OP to get confused when it comes to logs

e.g. ln2x => 1/2x or 3ln2x => 3/2x etc
They still give the right answer if you do it properly though. Like , etc... but yes, people do often get confused by differentiating logarithms.

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