Product Rule and Multiplying Factors

[Silvanoshei]

Sigh, hw is throwing me too many bones tonight.

y=(3-x²)(x ³-x+1)

We have our u and v, so, using the product rule, I'm setting that up as:

dy/dx = (3-x²)*d/dx(x³-x+1)+(x³-x+1)*d/dx(3-x²)

= (3-x²)(3x²-1+0)+(x³-x+1)(0-2x)

= (9x²-3-3x²+x²)+(????) ---> not sure where to go with this, did I mess up somewhere?

 

[DrPhil]

Sigh, hw is throwing me too many bones tonight.

y=(3-x²)(x ³-x+1)

We have our u and v, so, using the product rule, I'm setting that up as:

dy/dx = (3-x²)*d/dx(x³-x+1)+(x³-x+1)*d/dx(3-x²)

= (3-x²)(3x²-1+0)+(x³-x+1)(0-2x) <-- OK this far

= (9x²-3-3x²+x²)+(????) ---> not sure where to go with this, did I mess up somewhere?

Expand the multiplications and gather like powers of x.
What happened to the x^4 terms that must be in the answer?

Another approach that you could to to check your work:
Multiply out the two polynomials BEFORE differentiating.

 

[JeffM]

Sigh, hw is throwing me too many bones tonight.

y=(3-x²)(x ³-x+1)

We have our u and v, so, using the product rule, I'm setting that up as:

dy/dx = (3-x²)*d/dx(x³-x+1)+(x³-x+1)*d/dx(3-x²)

= (3-x²)(3x²-1+0)+(x³-x+1)(0-2x) Doing fine to here

= (9x²-3-3x²+x²)+(????) ---> not sure where to go with this, did I mess up somewhere?

\(\displaystyle \dfrac{dy}{dx} = (3- x^2)(3x^2 - 1+0)+(x^3 - x+1)(0-2x) = -(x^2 - 3)(3x^2 - 1) - (2x)(x^3 - x + 1) \implies\)

\(\displaystyle \dfrac{dy}{dx} = -(3x^4 - x^2 - 9x^2 + 3 + 2x^4 - 2x^2 + 2x) = -(5x^4 -12x^2 + 2x + 3) = - 5x^4 + 12x^2 - 2x - 3.\)

Let's check

\(\displaystyle y = (3 - x^2)(x^3 - x + 1) = 3x^3 - 3x + 3 - x^5 + x^3 - x^2 = - x^5 + 4x^3 - x^2 - 3x + 3 \implies\)

\(\displaystyle \dfrac{dy}{dx} = -5x^4 + 12x^2 - 2x - 3.\)

 

[Silvanoshei]

Thanks for the clarification JeffM! Made my nightmare HW into a clear breeze. *bow*