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C3 derivatives of trig functions

Hi,

I've got some questions and I'm not sure what to do.

An example of one is:

'By writing y=secx as y= (cosx)1(cosx)^{-1} use the chain rule to establish the result d(secx)dx=secxtanx \frac{d(secx)}{dx} = secxtanx

So after I've done the first bit, to get (cosx)1d(secx)dxsinx(cosx)2 (cosx)^{-1} \rightarrow \frac{d(secx)}{dx} sinx(cosx)^{-2} I'm not sure how to move on. I've got a number of questions that follow a similar pattern so hopefully if I can get a few tips with this one it will help me with the rest!

Thanks :rolleyes:
Original post by marcsaccount
Hi,

I've got some questions and I'm not sure what to do.

An example of one is:

'By writing y=secx as y= (cosx)1(cosx)^{-1} use the chain rule to establish the result d(secx)dx=secxtanx \frac{d(secx)}{dx} = secxtanx

So after I've done the first bit, to get (cosx)1d(secx)dxsinx(cosx)2 (cosx)^{-1} \rightarrow \frac{d(secx)}{dx} sinx(cosx)^{-2} I'm not sure how to move on. I've got a number of questions that follow a similar pattern so hopefully if I can get a few tips with this one it will help me with the rest!

Thanks :rolleyes:


That's correct....


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Reply 2
Original post by LightBlueSoldier
That's correct....


Posted from TSR Mobile


...but how do I arrive at the answer from my working so far?
Original post by marcsaccount
...but how do I arrive at the answer from my working so far?


Tan= sin/cos... Sec =1/cos


Posted from TSR Mobile
Reply 4
Original post by marcsaccount
...but how do I arrive at the answer from my working so far?




try writing the Cosx^-2 as secx^-2 :biggrin:

Then just make Sin/Cosx =tanx
Reply 6
vhvjhdbhfh, Hazard17 & LightBlueSoldier - thanks! I can worth through them all now :biggrin:

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