differentiating trig functions question (C4)

Original post by Noodle-Chan

So the question is: differentiate y=sin^2x/cos^3x

The workings I've got so far:

u=sin^2x
v=cos^3x

du/dx = 2sinxcos
dv/dx = -3sinxcos^2x

then I did the quotient rule and I got this in the end after simplifying:

dy/dx = sinx(2+3sin^2x)/cos^5x

but the answer says it's dy/dx = sinx(2+sin^2x)/cos^4x

so yeah... I kinda need help please, can someone explain where I went wrong? xD

Can you please show ALL your working out in order to determine where you went wrong? From the looks of it you simplified wrong.

Reply 3

OP

Ah yeah. Should have done that in the first place xD

ImageUploadedByStudent Room1448135697.406598.jpg

Posted from TSR Mobile

Reply 4

Original post by Noodle-Chan

Ah yeah. Should have done that in the first place xD

ImageUploadedByStudent Room1448135697.406598.jpg

Posted from TSR Mobile

On your second to last line you forgot to factorise (cosx)^2 on the numerator, which would be divided by the (cosx)^6 on the denominator. So that will leave you with (Cosx)^4 on the denominator.

*super big hint removed* now try to get the numerator in terms of sinx only.

(edited 8 years ago)

Reply 5

Ayman!

15

I'd rewrite your function to

Unparseable latex formula:

\[\frac{\mathrm{d} }{\mathrm{d} x}(\sec x\cdot tan^{2}x)\]

, it makes it much cleaner via the product rule in my opinion.

Reply 6

jamestg

Y13s - is this topic nasty? I see people asking for help with this topic EVERYWHERE.

Reply 7

Ayman!

15

Original post by jamestg

Y13s - is this topic nasty? I see people asking for help with this topic EVERYWHERE.

Not really. The poster has applied the chain rule and quotient rule well here but probably lost his way due to a silly mistake.

Reply 8

Original post by jamestg

Y13s - is this topic nasty? I see people asking for help with this topic EVERYWHERE.

Not really, it's just really easy to make silly mistakes and forget to simplify in earlier stages (or oversimplifying)

Reply 9

OP

Original post by kkboyk

On your second to last line you forgot to factorise (cosx)^2 on the numerator, which would be divided by the (cosx)^6 on the denominator. So that will leave you with (Cosx)^4 on the denominator.

*super big hint removed* now try to get the numerator in terms of sinx only.

I feel really really stupid asking this, but where's the cos^6(x) coming from? :s-smilie:

Reply 10

Original post by Noodle-Chan

I feel really really stupid asking this, but where's the cos^6(x) coming from? :s-smilie:

You're not stupid, its good to ask questions regardless if others find it stupid. You have to square the denominator from the original function when using the quotient rule. :tongue:

Reply 11

OP

Original post by kkboyk

You're not stupid, its good to ask questions regardless if others find it stupid. You have to square the denominator from the original function when using the quotient rule. :tongue:

Ahhh. I think I squared the denominator wrong using the quotient rule then? I squared cos^3(x) and I got cos^9(x). Is it meant to be cos^6(x) instead? :s-smilie:

Reply 12

Original post by Noodle-Chan

Ahhh. I think I squared the denominator wrong using the quotient rule then? I squared cos^3(x) and I got cos^9(x). Is it meant to be cos^6(x) instead? :s-smilie:

Yes

((cosx)^3)^2 = (cos x cosx cosx) (cosx cosx cosx) and using the rule of indices it becomes (cosx)^6 :tongue:

Edit: you just times the power together so 3x2 = 6, not 9.

Reply 13

OP

Original post by kkboyk

Yes

((cosx)^3)^2 = (cos x cosx cosx) (cosx cosx cosx) and using the rule of indices it becomes (cosx)^6 :tongue:

Edit: you just times the power together so 3x2 = 6, not 9

Ah! I think it was the trig functions incorporated in it that got me confused. xD
Thank you for your patience! I think I can work my way onward from here unless something else pops up. Thanks again! :biggrin:

Reply 14