The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Power Series.

Scroll to see replies

Tallon
oh God... you're right. I feel so stupid.
but did I get dv/dx and du/dx correct?
They look correct to me!
Reply 21
Avatar for Tallon
Tallon
OP
16
Phugoid
It seems you are a bit shaky on the chain rule/product rule.

What you are trying to differentiate are a product of functions of functions. i.e. f(g(x))×h(k(x)) f(g(x)) \times h(k(x)) .

To differentiate these, use the produce rule FIRST, then the chain rule on the two separate terms that pop out:

(f′(g(x))×g′(x))×(h(k(x))+f(g(x))×(h′(k(x))×k′(x)) \big(f'(g(x)) \times g'(x)\big)\times(h(k(x)) + f(g(x)) \times \big(h'(k(x)) \times k'(x)\big)

For functions that look like yours, it's this:

ddxAeBxcos⁡(Cx)=ABeBxcos⁡(Cx)−ACeBxsin⁡(Cx) \frac{d}{dx} Ae^{Bx}\cos(Cx) = ABe^{Bx}\cos(Cx) - AC e^{Bx} \sin(Cx)



Well, you're right. I am. It's just so ridiculous for a school to set this kind of homework for further maths, having not done this differentiating chapter in core maths c3 which this homework relies on.
Reply 22
Avatar for Tallon
Tallon
OP
16
DFranklin
They look correct to me!


so f''(x) = 9e^(3x)*cos(2x) - 6e^(3x)*sin(2x) - 4e^(3x)*cos(2x) - 6e^(3x)*sin(2x)

= 5e^(3x)*cos(2x) - 12e^(3x)*sin(2x)
= 5f(x) - 12e&(3x)*sin(2x)

not quite sure where the factor of f'(x) comes in, though.
unless, could you write it as 6f'(x) - -13f(x)?

whats the point of writing it in that form, though?
Tallon
so f''(x) = 9e^(3x)*cos(2x) - 6e^(3x)*sin(2x) - 4e^(3x)*cos(2x) - 6e^(3x)*sin(2x)

= 5e^(3x)*cos(2x) - 12e^(3x)*sin(2x)
= 5f(x) - 12e&(3x)*sin(2x)

not quite sure where the factor of f'(x) comes in, though.
unless, could you write it as 6f'(x) - -13f(x)?
Something like that yes (I haven't checked you've got the numbers right).

whats the point of writing it in that form, though?
Well, if you know that f''(x) = 6f'(x)+13f(x), then what's f'''(x)?

Spoiler



And you can keep doing that. So at this point you can keep taking derivatives without doing very much work. (Particularly if you just want to evaluate the derivatives when x = 0).

If you know what a Maclauren series is, you should see how this helps you calculate it.

Related discussions

Latest

Trending

Trending

Articles for you