Lately, my Maths teacher is talking about the application of the higher derivatives. To find the 2nd derivative and above is very tedious and a small mistake can screw the whole problem up especially those like eg : y = ((cos x)^0.5 )/ (x^3 + sec x)^3 Find the 2nd derivative.

Thus, I have been working to devise a shortcut general formula for nth derivative. And at last I found one possible way.......

First, break the expression into 2 function.

Hence, let y = f(x) g(x)

dy/dx = f' (x) g(x) + g' (x) f (x)
(d^2)y)/(dx)^2 = f'' (x) g(x) + f' (x) g' (x) + g''(x) f (x) + g' (x) f'(x)
= f'' (x) g(x) + 2 f' (x) g' (x) + g''(x) f(x)
(d^3)y)/(dx)^3 = f''' (x) g(x) + 3 f'' (x) g' (x) + 3 f' (x) g''(x) + g''' (x) f (x)

Hence, the differentiation of higher derivatives can be get through binomial expansion.....

Cool Right?

Hence, can some pros out there tell me other general formula for getting nth derivatives. Thanks.

Also, someone plz tell me whether this formula is fully optimized or not?