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  1. tom.evans.15

    Differentiation

    Just dont get confused between \sin^{-1}(x) and \frac{1}{\sin(x)}, they mean completely different things, \sin^{-1}(x) is the inverse function, and \frac{1}{\sin(x)} is the reciprocal of the function
  2. tom.evans.15

    Differentiation

    I would use the identity \frac{d}{dx}cosec(x)=\frac{d}{dx}\frac{1}{\sin(x)} and then do \frac{d}{dx}(\sin(x))^{-1} and use the chain rule to work it out... Similiarly I would use the identity sec(x)=\frac{1}{\cos(x)} to do the same hope i helped
  3. tom.evans.15

    Differentiating harder inverse trig

    Oh well. Doesn't matter... tbh, the explanation was more for my own sake rather than OPs or anyone elses... but i'm right, so i feel good now
  4. tom.evans.15

    Differentiating harder inverse trig

    Correct me if im wrong, but it IS the chain rule right? y=asin(f(x)) is in the form y=g(f(x)) where g(x) = asinx therefore dy/dx = g'(f(x)) * f'(x) so d/dx(asinf(x)) = (1/sqrt(1-(f(x)^2))*(f'(x)) and please, correct me if im wrong
  5. tom.evans.15

    Induction Statement

    What would be the best induction statement to use (ie the shortest and easiest to remember)? I've noticed some people say "As it is true for n=1, it is true for n=2, as it is true for n=2, it is true for n=3, therefore it is true for all real n>1" or something similiar, and I've also noticed...
  6. tom.evans.15

    binomial theorem.. in the expansion of... the coefficients bla bla bla

    I just tried that question, had difficulty... I think I compared coeffs wrong...
  7. tom.evans.15

    General Solutions to sinx and cosx

    You don't really need to derive them, but try choosing a set of integers (say 0,1,2) to use as n and sub it in, it gives you an idea of how the whole thing works. ie. when sinx=siny, x=n(pi)+((-1)^n)y for n=0, x=y (first quadrant) for n=1, x=pi-y (second quadrant) for n=2, x=2pi-y (third...
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