Recent content by MrSparky

Well basically to find the derivative its (derivative of original equation) . (power of original equation) . ( original equation)^(power of original equation - 1), in this case derivative of original equation is 4x^3 , power of original equation is 5, (original equation^ power of original...

dy/dx = gradient of tangent = 2-3x squared Then sub in point (1,2) and the gradient is -1 y=mx+b we know m = -1 y= -x + b Because the point (1,2) satisfy both the tangent and curve you sub it in 2= -1 + b b= 3 therefore the equation of the tangent is y= -x + 3

dy/dx = 2 sin x . -sin x +cos x . cos x= 2 ( cos squared - sin squared) So yes your correct

make the equation as x^2 - x^1 - 2 x^0 Now dy/dx = 2x - (1 . x^1) - (0 . 2x^-1) now after multiplying everything you get 2x - 1 - 0= 2x - 1 PS . The dots are multiplication signs as well if you didnt know that

oops, guess i misread that part when working it out, but yea your method is most likely better than mine.

Im not sure if this is the easiest way to do it but u can manipulate the equation to ((x-3)/(x^2 +1))+ (2/x-1) Then u split it up into a 3rd fraction so what you get is (x/x^2 +1) -(3/x^2 +1) + (2/x-1) dx then what u integrate i think u should get 1/2 ln (x^2 +1) - 1/3 ln (x^2 +1) + 2 ln...

lol guys if his asking a basic question like this i don't think it requires integration cos 2x = cos (x+x) cos (x+x) = cos x . cos x - sin x . sin x = cos^2 x - sin^2 x using the identity cos^2 x + sin^2 x = 1 sin^2 x= 1-cos^2 x cos^2 x +cos^2 x - 1 = cos 2x 2 cos^2 x -1=cos 2x There no...

sub u= sin t du/dt = cos t -250 sin^2 t . du/dt . dt = -250 u^2 du and then = (-250 u ^3)/3 Substitute 'u' back in and u get (-250 sin^3 t)/3 then the result is (-250 sin^3 t)/3 + c since there are no boundaries