Differnetiation

• Jan 1st 2013, 03:55 AM
Tutu
Differnetiation
Hi I just finished some questions and got these wrong, I won't get to ask anyone till a few days late, so I would really appreciate your help!

1.) Differentiate $\displaystyle \frac{1}{\sin\2\theta}$
Using quotient rule, I got $\displaystyle \frac{-2\cos\2\theta}{\sin^2\2\theta}$
Simplifying,
$\displaystyle -2(\cot\2\theta\ cosec\2\theta)$, why is this wrong?

2.) Differentiate $\displaystyle \tan^3(2x+1)$
I got $\displaystyle 6\tan^2\(2x+1)\sec^2\(2x+1)$, why is this wrong?

3.) If a, b and c are functions such that a(x) are not 0 for all x and c(x) = a(x)b(x) then c'(d)=0 for d is all real numbers such that a'(d)=0 or b'(d) =0.
I wrote that it's true becuase I thought the last part of the question made sense..but it's false, why is it so?

4.) Suppose a,b and c are functions such that a(x) is not 0 for all x and c(x) = b(a(x)) then c'(d)=0 for d is all real numbers such that a'(d)=0.
I wasn't sure so I thought it was false..

Thank you!
• Jan 1st 2013, 04:59 AM
abender
Re: Differnetiation
Quote:

Originally Posted by Tutu
Hi I just finished some questions and got these wrong, I won't get to ask anyone till a few days late, so I would really appreciate your help!
1.) Differentiate $\displaystyle \frac{1}{\sin\2\theta}$
Using quotient rule, I got $\displaystyle \frac{-2\cos\2\theta}{\sin^2\2\theta}$
Simplifying,
$\displaystyle -2(\cot\2\theta\ cosec\2\theta)$, why is this wrong?

Your answer here is correct. Perhaps your instructor wanted you to further simplify the expression using double-angle formulae? Still, I believe that's overkill. Why do you believe your solution is incorrect?
• Jan 1st 2013, 05:05 AM
abender
Re: Differnetiation
Quote:

Originally Posted by Tutu
2.) Differentiate $\displaystyle \tan^3(2x+1)$
I got $\displaystyle 6\tan^2\(2x+1)\sec^2\(2x+1)$, why is this wrong?

Again, why do you say you are wrong? Your answer here is correct, as well.
Happy New Year!