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thanks. corrected it. haha

lol so I start with solving it \frac{\partial f}{\partial x}=\frac{2x}{x^2+y^2}+2\frac{-yx^{-2}}{1+\frac{y^2}{x^2}}=\frac{2x}{x^2+y^2}-\frac{2y}{x^2+y^2} \frac{\partial^2 f}{\partial x^2}=\frac{2(x^2+y^2)-(2x-2y)(2x)}{(x^2+y^2)^2}=\frac{-2x^2+2y^2+4xy}{(x^2+y^2)^2} \frac{\partial...

This one is just simple straight forward partial derivatives. Try this: If f is differentiable, find the limit \lim_{h\rightarrow0}\frac{f^2(x+3h)-f^2(x-h)}{h}

These explainations make the question make more sense. lol I was trying to say the question might be solve or find the ananytic solution of y^\prime+y^2=y\cos x

for question 1: so it was not "find y", it was the approximation solution by series. that's totally different. if the equation was correct, no way to find the analytic solution! for question 2: yes my substitution works and is correct. as a matter of fact, only when it is a factor, i can do...

are you sure the first question is really like that? I guess you missed a y on RHS? like y^\prime+y^2=y\cos x the second question can be solved, other than L'Hospital rule, using equivalent substitution like c^{\frac1n}-1\sim \ln c^{\frac1n}=\frac1n\ln c

Re: HSC 2014 4U Marathon Apply the Cosine rule once in each of triangles ABX and ACX, then multiply the first equation by n second by m and add up.

Re: HSC 2014 4U Marathon - Advanced Level so confusing a question. I believe \theta should not be constant, is that correct? and launching from the equator, isn't the gravity pointing to the centre of the planet? if yes, it is not downward. or maybe I totally understand the question wrong?

Re: MX2 Integration Marathon lol and this \int\frac{1}{x\ln x\ln\ln x\ln\ln\ln x}dx

Re: HSC 2014 4U Marathon - Advanced Level wow, very nice, especially the way you prove continuous C-S inequality by using discrete C-S. I would use positive definite quadratic and negative discrimint

Re: MX2 Integration Marathon lol e^{x+e^x}=e^x\cdot e^{e^x}=\frac{d}{dx}e^{e^x}

Re: HSC 2014 4U Marathon - Advanced Level quite so. and I was trying to say similarly prove that \left[\int_a^bf(x)g(x)dx\right]^2\leq\int_a^bf^2(x)dx\cdot\int_a^bg^2(x)dx where a<b .

Re: HSC 2014 4U Marathon - Advanced Level Show that for any real numbers a_i, b_i, i=1,2,\cdots,n , where n is a positive integer, \left(\sum_{k=1}^na_kb_k\right)^2\leq\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)

Re: MX2 Integration Marathon not sure this one posted before? \int\frac{\tan x+\tan 2x+\tan 3x}{\tan x\tan 2x\tan 3x}dx

Re: MX2 Integration Marathon hah, this is my favorite one. use substitution x=\frac{\pi}{4}-u then notice that \ln\left[1+\tan\left(\frac\pi4-u\right)\right]=\ln\frac2{1+\tan u}=\ln 2-\ln(1+\tan u) It follows that I=\int_0^{\frac\pi4}\ln2dx-I which gives I=\frac{\pi}{8}\ln2

Re: MX2 Integration Marathon ok I reckon it is not a good question coz I hate partial fractions too. I wrote this question was to try using the technique x^4=x^4-x+x which reduce the integral into \frac{x}{x^3+1} via cancelling common factor, and \frac{1}{u^3-1} via substitution...

Re: MX2 Integration Marathon Dont forget to try this one

Re: HSC 2014 4U Marathon - Advanced Level A method beyond HSC scope is: Let f(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}$ . Find f^\prime(x) , then integrate. Finally the desired sum is -f(-1) . Alert: 1. sum of (functional) series 2. termwise differentiation applies 3. f(x) only...

Re: MX2 Integration Marathon \int\frac{x^4}{x^6-1}dx

Re: HSC 2014 4U Marathon - Advanced Level Well, a hard one, belongs to Olympiad question not HSC, although the proof is only a few lines. Here is the outline of proof, to fully understand, you need draw a diagram. Let feet of perpendicular on sides AB, AC are respectively D and E. Let FH...