Consider e^(Cos(x))^4 thats e to cos x to the 4th btw
Differentiate completely:
EDIT: also could you NOT do it in dy/dx du/dx form cuz that just confuses. me lol.....
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Rev0lution
United States1805 Posts
Consider e^(Cos(x))^4 thats e to cos x to the 4th btw Differentiate completely: EDIT: also could you NOT do it in dy/dx du/dx form cuz that just confuses. me lol..... | ||
kpcrew
Korea (South)1071 Posts
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micronesia
United States24345 Posts
On October 16 2008 23:46 kpcrew wrote: could have sworn it was outer * derivative of inner + inner * derivative of outer This looks like the product rule which is not what this is. More like the chain rule. do you mean [e^(cos(x)]^4 | ||
Rev0lution
United States1805 Posts
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micronesia
United States24345 Posts
in this case df/dx is -sin(x)*e^cos(x) | ||
Rev0lution
United States1805 Posts
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micronesia
United States24345 Posts
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BlackStar
Netherlands3029 Posts
Derivative of [u(x)]^4 is 4[u(x)]^3 times the derivative of u(x) So: f(x)=[e^(cos(x)]^4 f'(x)=4[e^(cos(x)]^3 * [e^(cos(x)]' f'(x)=4[e^(cos(x)]^3 * -sin(x)[e^(cos(x)] Be sure you remember what is the derivative of e^u(x). No need to be confused by dy/dx = dy/du * du/dx and stuff like that. Know the derivative operators. No real math involved here. Just remember the symbols eventhough there are several with the same meaning. could have sworn it was outer * derivative of inner + inner * derivative of outer Yes, product rule confusing. (f*g)' = f*g' + f'*g | ||
inlagdsil
Canada957 Posts
On October 16 2008 23:45 Rev0lution wrote: I forgot how to do this and I dont have my calc book with me if anyone can give me a bump in the right direction?? Consider e^(Cos(x))^4 thats e to cos x to the 4th btw Differentiate completely: EDIT: also could you NOT do it in dy/dx du/dx form cuz that just confuses. me lol..... In the beginning I found dy/dx form confusing but as you go on in math it will make sense and in fact be less confusing. Example: f = 35x+27y-3z, WTF is f ' ?? | ||
Mastermind
Canada7096 Posts
On October 17 2008 04:11 inlagdsil wrote: Show nested quote + On October 16 2008 23:45 Rev0lution wrote: I forgot how to do this and I dont have my calc book with me if anyone can give me a bump in the right direction?? Consider e^(Cos(x))^4 thats e to cos x to the 4th btw Differentiate completely: EDIT: also could you NOT do it in dy/dx du/dx form cuz that just confuses. me lol..... In the beginning I found dy/dx form confusing but as you go on in math it will make sense and in fact be less confusing. Example: f = 35x+27y-3z, WTF is f ' ?? In that case it doesnt make sense to ask what f' is. There is no single derivative. You could ask what is the partial derivative with respect to x or y or z, but to just ask for a derivative makes no sense. Unless of course 2 of those symbols are actually constants and not variables, but I assume you meant for all 3 to be variables. | ||
HeavOnEarth
United States7087 Posts
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Ecael
United States6703 Posts
On October 18 2008 03:17 HeavOnEarth wrote: the algebra in calculus is harder than the actual calculus imo =/ Amen to that, addition and division is way harder than calculus. At times I wonder why we teach math the way we do. | ||
HeavOnEarth
United States7087 Posts
On October 18 2008 05:31 Ecael wrote: Show nested quote + On October 18 2008 03:17 HeavOnEarth wrote: the algebra in calculus is harder than the actual calculus imo =/ Amen to that, addition and division is way harder than calculus. At times I wonder why we teach math the way we do. usa is quickly falling so far behind in math and science | ||
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