How do you differentiate #y=(x^5+1)(3x^2-x)#?

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Answer 1 · 2016-12-07T07:21:39

The answer is #dy/dx=21x^6-6x^5+6x-1#

We use #(x^n)'=nx^(n-1)#

You apply the product role of differentiation

#(uv)'=u'v+uv'#

Here,

#u=(x^5+1)#, #=>#, #u'=5x^4#

#v=(3x^2-x)#, #=>#, #v'=6x-1#

#dy/dx=5x^4(3x^2-x)+(x^5+1)(6x-1)#

#=15x^6-5x^5+6x^6-x^5+6x-1#

#=21x^6-6x^5+6x-1#