How do you find the derivative of #y= (1/3)x^3 - 5x - 4/x#?

1 Answer
Apr 10, 2015

If you really want to do this using first principles (the definition), I'll go through it, but You'll need to ask for that specifically, because it's a lot of typing for me.

Using the power rule and the other properties of derivatives, we do the following:

#y= (1/3)x^3 - 5x - 4/x = (1/3)x^3 - 5x - 4x^(-1)#, So

#y'= (1/3)[3x^2] - 5 - 4[-1x^(-2)] = x^2-5+4/x^2#

That's all the calculus. If desired, we could do the algebra to get a common denominator:

#y' = (x^4-5x^2+4)/x^2#

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