The antiderivative of the following functions are sought:
1)(x)^1/2. The solution is, in my booklet, 1/2(x)^1/2
But if you use the quotient rule ( f(x)/g(x)' => f'(x)g(x)-f(x)g'(x)/ (g(x))^2)
you get 1/-4x(x)^1/2. Where's the problem? I know, that if you consider 1/(x)^1/2 as x^-1/2 it works out, i just dont see the mistake with applying the quotient rule
2) 2(x^3)^1/2 Here I'm clueless, how do you find the antiderivative at this example?
3) 3sin(x). to me the solution is 3(-cos(x)). In my booklet the solution is 3(cos(x)), but isn't the derivative of cos -> -sinus?
Thanks for you help
Sorry I dount understand anything of this.
Originally Posted by harish21
1)Is as antidervative of henceforth wrong?
2)why does the quotient rule appereantly not apply here or made i a mistake?
3) what about 2,3?
sry i cannot edit the last post, please ingore it!
Thanks for your help!