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Hi,
I need to find a Derivative of a drigonometric function.
1. F(x)= sin x sec x
That how i did it: f'(x)= Cos x Sec x Sec x Tan x Sin
But the answer shows f'(x)= Sec² x
Question: Is mine wrong or is the answer wrong? Please explain how to do it.
Thank you)
Also, how to solve this type of derivative
1/Sec x Tan X
Can anybody solve it for me, please and explain.
Thank you
Last edited by Gercel (2009-07-05 04:06:22)
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Hi Gercel;
For #1;
The book answer is right, here's why:
By the product rule:
Now remember:
Last edited by bobbym (2009-07-05 04:57:09)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hmm, what about
Sinx Secx
Sin x will be u u'v+uv'
Sec x will be v Cos x (Sec x) + Sin x (Sec x Tanx)
After that I dont know how yo simplified to Tan x^2 +1 and got sec ^2 x
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Hi Gercel;
is one that I just know. But it is easy to prove.
Last edited by bobbym (2009-07-05 05:06:10)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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One more question, when I do derivatives do I always have to put d/dx, dx/dy or I can simply put f '(x) in front and do the derivative without d/dx or dy/dx
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Hi Gercel;
Yes, f'(x) is the same as dy/dx.
Here's how to use the trig subs on problem #1:
OK, what else is there? I am losing my way.
Last edited by bobbym (2009-07-05 05:25:37)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I kind of understood it.)) Thankx
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