1. ## Differentiating Sin Cos

Help!! Just dabbling into differentiation...

f(x) = cos (3x)

f'(x) = -sin (3x)

RIght?

2. You need to use the chain rule to do this.
until your good with the rule do it manually using the formula below.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

so let $u = 3x$
then you got $y = \cos u$

then use the chain rule to find $\frac{dy}{dx}$

3. I think I got this one right?

k(x) = x^7 cos(4x)

f(x) = x^2
f ' (x) = 2x

g(x) = cos(4x)
g ' (x) = -4sin(4x)

by the product rule:

k'(x) = f '(x) g (x) + f (x) g'(x)
= (2x) (cos (4x) + x^2 (-4sin(4x))

I can't simplify that anymore, right?

4. Originally Posted by Sweeties
I think I got this one right?

k(x) = x^7 cos(4x)

f(x) = x^2
f ' (x) = 2x

g(x) = cos(4x)
g ' (x) = -4sin(4x)

by the product rule:

k'(x) = f '(x) g (x) + f (x) g'(x)
= (2x) (cos (4x) + x^2 (-4sin(4x))

I can't simplify that anymore, right?

if you wrote
$2x( \cos 4x - 4x^2 \sin 4x )$ which is what i am reading your wrong.

but I am fairly sure you meant.

$2x \cos 4x - 4x^2 \sin 4x$ which is correct, good stuff.

5. I thik my basic maths skills let me down more than the ability to grasp higher concepts

I have differentiated a question using quotient rule and got as far as:

k'(x) = ((4-t^2) (3e^(3t)) - (e^(3t)) (-2t)) / (4-t^2)^2

I put it into QuickMath but it comes up all funny

so, what next? can i simplify this further, if so how? thanks.

6. I have no idea what the question is, but from your answer i guess you have been told to differentiate $k(t) = \frac{ e^{3t}}{4-t^2}$

I am finding it difficult to read your answer, (maybe you should try using LaTeX) but it looks like you have written. $k'(t) = \frac{ 3e^{3t}(4-t^2) + 2t \cdot e^{3t}}{(4-t^2)^2}$
which is correct.

7. Originally Posted by bobak
I have no idea what the question is, but from your answer i guess you have been told to differentiate $k(t) = \frac{ e^{3t}}{4-t^2}$

I am finding it difficult to read your answer, (maybe you should try using LaTeX) but it looks like you have written. $k'(t) = \frac{ 3e^{3t}(4-t^2) + 2t \cdot e^{3t}}{(4-t^2)^2}$
which is correct.
Yes, that's the question. Sorry, I should have written it out in teh first place.
I have Googled latex and will use it in future.

So the answer can't be simplified further?

8. well you can take out a factor of $e^{3t}$ the numerator. Can't see anything else.

9. ..

10. ...sorry. just getting used to all this!

11. I hope you didn't pay for QuickMath, because it is wrong.

$ln( 3 \sin (6x) ) = \ln (3) + \ln ( \sin (6x) )$

$\frac{d}{dx} \left ( \ln (3) + \ln ( \sin (6x) ) \right ) = 0 + \frac{6 \cos (6x) }{ \sin (6x)} = 6 \cot (6x)$

12. Originally Posted by bobak
I hope you didn't pay for QuickMath, because it is wrong.

$ln( 3 \sin (6x) ) = \ln (3) + \ln ( \sin (6x) )$

$\frac{d}{dx} \left ( \ln (3) + \ln ( \sin (6x) ) \right ) = 0 + \frac{6 \cos (6x) }{ \sin (6x)} = 6 \cot (6x)$
OK. Can you explain how you go did this step?

$ln( 3 \sin (6x) ) = \ln (3) + \ln ( \sin (6x) )$

-Don't forget, I'm new to all this! Be gentle please!

13. you use the laws if logs. Are you fimmilar with them ? ( you should have been taught them in your C2 math module).

$\log ab = \log a + \log b$

14. Originally Posted by bobak
you use the laws if logs. Are you fimmilar with them ? ( you should have been taught them in your C2 math module).

$\log ab = \log a + \log b$
Oh yes! I forgot. Thank you.

Also, you have used the chain rule in your solution and I need to use the composite rule:
k'(x)=g'(f(x))f'(x)

which I think is more complicated, and I can't seem to get the same answer as you got. I will have another go now that you've reminded me of the log rule

15. Originally Posted by Sweeties
Oh yes! I forgot. Thank you.

Also, you have used the chain rule in your solution and I need to use the composite rule:
k'(x)=g'(f(x))f'(x)

which I think is more complicated, and I can't seem to get the same answer as you got. I will have another go now that you've reminded me of the log rule

One think that you need to realise that all the rules different repertations of the chain rules are infact the same.

I advise you use the chain rule as $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
$
and with you will be able to do this quickly and accurately.

so in this case we have $y = \ln ( \sin (6x) )$

$u = \sin (6x)$
$\frac{du}{dx} = 6 \cos (6x)$
$y = \ln u$
find dy/du

then finish it off