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,

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

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

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

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

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

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