# Math Help - General differentiation questions

1. ## General differentiation questions

Hi

Some questions I can't do. This one is Q6. The textbook answers are
x < - 2^0.5, x > 2^0.5

BTW I MEAN +/- ROOT 2 FOR THE LAST BIT OF WORKING

2. Originally Posted by Apex
Hi

Some questions I can't do. This one is Q6. The textbook answers are
x < - 2^0.5, x > 2^0.5

BTW I MEAN +/- ROOT 2 FOR THE LAST BIT OF WORKING
hint. a fraction is negative (that is less than zero) if we divide a negative number by a positive number, and vice versa (obviously, division by 0 is a no-no).

the denominator of f'(x) is a perfect square, thus it is always greater than or equal to zero. so to find where f'(x) < 0, we simply solve for where the numerator is < 0

so you want to solve $\left( x^2 + 2 \right) - 2x^2 < 0$

you ended up doing this, but your mistake was this: if $x^2 = 2$ then $x = \pm \sqrt{2} = \pm 2^{0.5}$

3. Hi

Is -2^0.5 equivalent to (-2)^0.5 :. wrong?

I don't see how to get to the final answer though. How do I know which way round the inequalities are, and whether the solutions are seperate from eachother?

Question 7 is below.

Thanks v much
Apex

4. Originally Posted by Apex
Hi

Is -2^0.5 equivalent to (-2)^0.5 :. wrong?
yes, that is wrong.

$-2^{0.5} = - \sqrt{2}$ while $(-2)^{0.5} = \sqrt{-2}$ ...which is not a real number

I don't see how to get to the final answer though. How do I know which way round the inequalities are, and whether the solutions are seperate from eachother?
draw a number line and place $\sqrt{2}$ and $- \sqrt{2}$ in their relative positions, then check numbers in the three regions in the formula for f'(x). so, for instance, you could try -2, 0 and 2. that way you can figure out what intervals work

5. Originally Posted by Apex
Question 7 is below.

Thanks v much
Apex
$8^{2/3} = 4$

6. Oh yeh thanks. I did it wrong way round. As for the numberline idea, I think there is a graphical method of doing this (how do my graphs help)? I need the graphical way if it's possible.

I've also tried Q8, which I get incorrect answers for.
Textbook answers are 0.25, 1.57, 2.89, 4.71.

Thanks so much!

7. Hello,

This is question 9. Textbook doesn't say anything about it (it usually leaves "prove" answers out).

I'll post more questions tomorrow, unless I can somehow manage to complete them all correctly :P

Thanks

8. Originally Posted by Apex
Oh yeh thanks. I did it wrong way round. As for the numberline idea, I think there is a graphical method of doing this (how do my graphs help)? I need the graphical way if it's possible.
the graphical way would be to graph $y = 2 - x^2$ and see the regions where y is less than zero (where the graph is under the x-axis, this is a lot more work than using the number line method however, usually, here, it's about the same, since 2 - x^2 is easy to graph)

9. Originally Posted by Apex
I've also tried Q8, which I get incorrect answers for.
Textbook answers are 0.25, 1.57, 2.89, 4.71.

Thanks so much!
in the line where you divided through by cos(x), don't. you may be dividing by zero (in fact, having cos(x) = 0 is one way to get the desired result).

$-2 \sin 2x + \cos x = 0$

$\Rightarrow - 4 \sin x \cos x + \cos x = 0$

$\Rightarrow \cos x ( -4 \sin x + 1 ) = 0$

$\Rightarrow \cos x = 0 \mbox { or } -4 \sin x + 1 = 0$

now continue

10. Originally Posted by Apex
Hello,

This is question 9. Textbook doesn't say anything about it (it usually leaves "prove" answers out).

I'll post more questions tomorrow, unless I can somehow manage to complete them all correctly :P

Thanks
you differentiated incorrectly

$y = x \sqrt{ \sin x }$

$\Rightarrow \frac {dy}{dx} = \sqrt{ \sin x } + \frac {x \cos x}{2 \sqrt{ \sin x}}$

(you forgot to use the chain rule when differentiating the $\sqrt{ \sin x}$)

now continue

11. Hi

I carried with Q9 but got stuck. See Q9 below.

Thanks alot
Apex

12. This one requires more of an explanation

13. Stuck on solving this....

14. I don't understand what this means.

15. bump

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