A nutter way....


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Posted by Subhotosh Khan on September 29, 2002 at 16:48:35:

In Reply to: Re: calculus! help! posted by Denis Borris on September 29, 2002 at 16:21:37:

: I'll do this one (hope I'm correct!)

: : Find the derivative if Y= (3x^5-7x+5)(x^3+x-a)

Y= (3x^5-7x+5)(x^3+x-a)

Y' = (15x^4-7)(x^3+x-a)+ (3x^5-7x+5)(3x^2+ 1)

and I'll leave it at that ....



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