Help!! Just dabbling into differentiation...
f(x) = cos (3x)
f'(x) = -sin (3x)
RIght?
You need to use the chain rule to do this.
until your good with the rule do it manually using the formula below.
$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
so let $\displaystyle u = 3x$
then you got $\displaystyle y = \cos u$
then use the chain rule to find $\displaystyle \frac{dy}{dx} $
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.
I have no idea what the question is, but from your answer i guess you have been told to differentiate $\displaystyle k(t) = \frac{ e^{3t}}{4-t^2}$
I am finding it difficult to read your answer, (maybe you should try using LaTeX) but it looks like you have written. $\displaystyle k'(t) = \frac{ 3e^{3t}(4-t^2) + 2t \cdot e^{3t}}{(4-t^2)^2}$
which is correct.
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
I am guessing that your reading the rules out of your C3 book.
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 $\displaystyle \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 $\displaystyle y = \ln ( \sin (6x) )$
$\displaystyle u = \sin (6x)$
$\displaystyle \frac{du}{dx} = 6 \cos (6x)$
$\displaystyle y = \ln u$
find dy/du
then finish it off