1123581321 Posted September 27, 2011 Share Posted September 27, 2011 I was wondering whether someone would be able to solve this equation by getting the maxima and minima and show the working: -3cos(4x-pi) I would attempt it, but i'm not quite sure how to go about it...thanks. Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted September 27, 2011 Share Posted September 27, 2011 (edited) This seems a lot like a homework question. We generally don't provide complete answers to such things. Perhaps you could make a start and we'll guide you through the rest. Some questions to which the answers should get you started: Do you know what the derivative of the cosine function is? Do you know what the chain rule is? Do you know what it means when the derivative of a function is zero If you have more general questions (ie. you need an explanation of the chain rule) feel free to ask. Edited September 27, 2011 by Schrödinger's hat Link to comment Share on other sites More sharing options...

1123581321 Posted September 27, 2011 Author Share Posted September 27, 2011 ok, sorry, i realise the homework section would have been more appropriate. I realise that you differentiate it first. but where do you go from there...? how exactly does u substitution leave you with 3x. and how do you differentiate pi ? Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted September 27, 2011 Share Posted September 27, 2011 (edited) I realise that you differentiate it first. but where do you go from there...? Well, when the derivative is zero you have an extreme of your function. Can you solve your derivative for zero? how exactly does u substitution leave you with 3x. and how do you differentiate pi ? Hmm, pi is a constant, it doesn't change. I'll give you another nudge along the way. We want: [math]\frac{d}{dx}-3\cos{(4x-\pi)}=0[/math] First, carry the constant out the front, it doesn't matter. We recognise we want to use the chain rule, so we can set [math]u=4x-\pi[/math] and identify the rest as it as: [math]-3 \frac{d}{du}\cos{u}\frac{du}{dx}=0[/math] (If you need more explanation of how/why this works, do ask) Splitting our problem into two problems we hopefully already know how to solve: First, the derivative of [math]\cos{u}[/math] with respect to u (you can safely ignore what u might depend on, and all the rest of the problem for this step) Second, the derivative of [math]4x-\pi[/math] with respect to x. One thing that might help is to remember that derivatives are linear, the derivative of the sum is the sum of the derivatives. Hopefully you can do: [math]\frac{d(4x)}{dx}[/math] easily enough. And you asked what the derivative of [math]\pi[/math] (with respect to x) is. Think about what 'derivative' means. How fast does the value of [math]\pi[/math] change as we change x? Edited September 27, 2011 by Schrödinger's hat Link to comment Share on other sites More sharing options...

1123581321 Posted September 28, 2011 Author Share Posted September 28, 2011 ok. and just another thing. when using/applying the product rule, what out of; addition, subtraction, multiplication & division can be used ? Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted September 29, 2011 Share Posted September 29, 2011 Uhmm, I'm not quite sure what you mean. The product rule is: [math] \frac{d}{dx} f(x)g(x) = \frac{d f(x)}{dx} g(x) + f(x)\frac{dg(x)}{dx}[/math] I can explain a bit about the intuition behind it if you like, or perhaps you could re-word your question? Link to comment Share on other sites More sharing options...

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