`y = x^2(x^3-2)` Find dy/dx two different ways. Distribute x^2 before differentiating. Use product rule on the two factors. Show the answers are equivalent.

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

`y = x^2(x^3-2)`

Solution I.

If we are going to distribute `x^2 ` to `x^3 - 2` first, we would have:

`y = x^5 - 2x^2`

Then, take the derivative of each term using power rule which is `(u^n)' = n*u^(n-1) *u'` .

`(dy)/(dx)= (x^5-2x^2)' = (x^5)' - (2x^2)'= 5x^4*x' - 2*2x*x'=5x^4*1 - 4x*1`

`(dy)/(dx) = 5x^4-4x`

Solution II.

If we are going to apply the product rule which is `(u*v) = uv' + vu'` ,  let:

`u = x^2 `                     and

`v = x^3 - 2`

Then, let's determine u' and v'. 

Using the power rule of derivative,

`u'= 2x*x'=2x*1=2x`   and

`v'=3x^2*x' -2'=3x^2*1-2' = 3x^2-2'`

Note that the derivative of a constant is zero ( c' = 0). So,

`v'=3x^2 - 0 = 3x^2`

Then, substitute u,v, u' and v' to the product formula of derivative.

`(dy)/(dx)= x^2(3x^2) + (x^3-2)(2x)`

And, simplify.

`(dy)/(dx)=3x^4 + 2x^4-4x=5x^4-4x`

This proves that either methods result to same derivative of the function.

Hence, the derivative of `y = x^2(x^3-2)`  is  `(dy)/(dx)=5x^4-4x` .

Sameera W's profile pic

Sameera W | (Level 1) eNoter

Posted on

y=x^2(x^3-2)

We can distribute X^2  before differentiating as follows

y= x^2 .x^3  - 2.x^2    (1) 

by simplfying (1)

y= X^5 -2.x^2              (2)

........................................................

remember the following basic rules

* Y=x^n then dy/dx =n.x^(n-1)

* Y= U.V (U and V are function of x)

   dY/dx =U.dV/dx+V.dU/dx

.........................................................

solution i.

from (1),

dy/dx=( x^2 . (3.x^2) +x^3.(2.x^1) ) -2.2.x^1

dy/dx=3.x^4+2.x^4 -4.x

dy/dx= 5x^4 - 4x (this is the answer)

solution ii.

from (2),

dy/dx =5.x^4-2.(2.x^1)

dy/dx = 5x^4 - 4x (this is the answer)

 

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