finding the derivative?

raven2k7

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so i am stuck at this point. according to the book the correct answer would be -2/(sex(x)-cos(x))^2. I cant figure out how they got the *2* at the top part. i have uploaded my working
 

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Would you please resend with the picture in a readable orientation.
 
hi, sorry , let me know if this is better.
The work so far is good, except that again you wrote f(x) for f'(x). They are not the same thing!

The answer you gave might be accepted by a teacher, but to get the book's answer, you need to simplify. Expand each square in the numerator, and cancel or combine many terms. Give it a try, and let us see what you get.
 
Yes, it is much better. Thank you.

Couple of points.

It is a formal error to define f(x) as a fraction and as the fraction's numerator or denominator. A quick glance did not reveal to me any mistake arising from that here, but it is likely to lead to mistakes.

Try [MATH]f(x) = \dfrac{g(x)}{h(x)}.[/MATH]
You seem to want us to compare the book's answer to yours, but you have not told us what the book's answer is.

I suggest you do the squaring and remember that

[MATH]sin^2(x) + cos^2(x) = 1.[/MATH]
 
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Yes, it is much better. Thank you.

Couple of points.

It is a formal error to define f(x) as a fraction and as the fraction's numerator or denominator. A quick glance did not reveal to me any mistake arising from that here, but it is likely to lead to mistakes.

Try [MATH]f(x) = \dfrac{g(x)}{h(x)}.[/MATH]
You seem to want us to compare the book's answer to yours, but you have not told us what the book's answer is.

I suggest you do the squaring and remember that

[MATH]sin^2(x) + cos^2(x) = 1.[/MATH]
I did give the books answer in my first message, but i will upload it in this reply. ok so i see what u wrote but what confuses me is where did u learn this rule from?[MATH]sin^2(x) + cos^2(x) = 1.[/MATH] ?
 

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The work so far is good, except that again you wrote f(x) for f'(x). They are not the same thing!

The answer you gave might be accepted by a teacher, but to get the book's answer, you need to simplify. Expand each square in the numerator, and cancel or combine many terms. Give it a try, and let us see what you get.
ok, ill give it a try . lets see
 
Actually, I should have written [MATH]sin^2(x) + cos^2(x) \equiv 1.[/MATH]
It is one of the basic identities in trigonometry.

It comes directly from the triangular definition of the trigonometric functions and the Pythagorean theorem.

Imagine a right triangle, where angle alpha is not the right angle, and let the side opposite to angle alpha be A, the side adjacent be B and the hypotenuse be C.

[MATH]A^2 + B^2 \equiv C^2 \implies \dfrac{A^2}{C^2} + \dfrac{B^2}{C^2} \equiv \dfrac{C^2}{C^2} \equiv 1 \implies[/MATH]
[MATH]\left ( \dfrac{A}{C} \right )^2 + \left ( \dfrac{B}{C} \right )^2 \equiv 1 \implies sin^2(\alpha) + cos^2(\alpha) \equiv 1.[/MATH]
 
Actually, I should have written [MATH]sin^2(x) + cos^2(x) \equiv 1.[/MATH]


No, it was sufficient as you had it back in post # 5 with the equals sign, because identities are equalities, though Wikipedia mentions that "sometimes" the equivalence symbol is used.

So, everywhere else in your post you can replace that equivalence symbol with
the equals sign.
 
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