Question

Differentiate 3x² over x³ +3

Answer 1

`\displaystyle{y' = \frac{ - 3x( {x}^(3) + 6) }{( {x}^(3) + 3)^(2) }}`

Explanation:

We can use the quotient rule:

`\displaystyle{y = (u)/(v) \to y '= \frac{u'v - uv'}{ {v}^(2) }}`

In this case, u and v both are functions. By applying formula above, we will have:

`\displaystyle{y' = \frac{(3 {x}^(2))' ( {x}^(3) + 3) - (3 {x}^(2) )( {x}^(3) + 3)' }{( {x}^(3) + 3)^(2) }}`

Then apply power rule to derive functions, we have to know that:

`\displaystyle{y = {x}^(n) \to y' = n {x}^(n - 1) } \\ \\ \displaystyle{y = c \to y' = 0 \: \: \: (c \in \mathbb{R})}`

Therefore, we will have:

`\displaystyle{y' = \frac{6x ( {x}^(3) + 3) - (3 {x}^(2) )3 {x}^(2) }{( {x}^(3) + 3)^(2) }}`

Expand the terms in:

`\displaystyle{y' = \frac{6 {x}^(4) + 18x - 9 {x}^(4) }{( {x}^(3) + 3)^(2) }} \\ \\ \displaystyle{y' = \frac{ - 3 {x}^(4) + 18x }{( {x}^(3) + 3)^(2) }} \\ \\ \displaystyle{y' = \frac{ - 3x( {x}^(3) + 6) }{( {x}^(3) + 3)^(2) }}`

Hence, that's the answer. You can keep in unfactored form or factored form as you desire!