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Hello. I have just uploded onto Photobucket 16 calculus practice problems, all of which will be of similar content to my eventual final exam. Now, an answer to these questions would be nice, but is not necessary. Instead, I am concerned with and interested in THE METHOD as to how one could arrive to the correct answer, as well as an EXPLANATION as to why each step needs to be performed in relation to the ones that follow. Anybody who can break down and explain the contents of these problems to me in a SIMPLISTIC way would just be amazing in my book. So thank you very much should you decide to help enlighten me, as I appreciate this more then the words in this post could even begin to convey!

Question #3: http://www.i138.photobucket.com/albums/q271/Link3324/math3.jpg


Full Album: http://www.s138.photobucket.com/albums/q271/Link3324/

2006-12-11 12:52:20 · 2 answers · asked by link332456 1 in Science & Mathematics Mathematics

No. This is NOT any kind of required assignment. it is a guide to work from and to understand before the actual test. Any help would be greatly appreciated, as I have used yahoo answers before and the people here are really good at breaking things down, even better then teachers can sometimes!

2006-12-11 12:57:40 · update #1

2 answers

check out http://www.schoolpiggyback.com ...u cna get ur entire math assignment done by other students....goodluck : )

2006-12-11 12:53:53 · answer #1 · answered by lori b 2 · 0 1

Problems of this sort are of the type f(g(x)), a function of a function, and they are solved in general in the following way (chain rule):

d(f(g))/dx = df/dg * dg/dx;

in your problem f(g) = e^g, and g(x) = 2tan√x

First take the derivative of f with respect to g to get e^g and multiply by dg/dx. So the first step gives the result

e^g * dg/dx

Now you need the derivative of g, and you do the same thing again. This time g(h) = 2tan(h) and h = √x. Chain becomes

e^x * d/dh 2tan(h) * d/dx √x

e^x * 2sec^2(x) * .5x^-.5

The idea is to find the functions within the functions, and treat the functions as variables to simplify.

2006-12-11 21:06:09 · answer #2 · answered by gp4rts 7 · 0 0

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