# Thread: A single quick differentation

1. ## A single quick differentation

Answer at the back of the book is different than what I come out. Im pretty sure there are different ways to get it, but If you would differentiate...
F(x)=3x^2 (5-x)^4

how would you get it to be f '(x)=(21x^2 - 14x)(x^3 -2x +1)^6 ?

tyvm!

2. Originally Posted by 3deltat
Answer at the back of the book is different than what I come out. Im pretty sure there are different ways to get it, but If you would differentiate...
F(x)=3x^2 (5-x)^4

how would you get it to be f '(x)=(21x^2 - 14x)(x^3 -2x +1)^6 ?

tyvm!
That solution is not possible. I went ahead and did some of the work (before noticing that the solution is not possible) so I'll post it anyways. However, the original function was a 6th degree polynomial (if you have multiplied it all out), but the solution you posted is a 20th degree polynomial. That's clearly wrong.

$f(x)=3x^2(5-x)^4$

$f'(x)=\frac{d}{dx}[3x^2]*(5-x)^4+3x^2\frac{d}{dx}[(5-x)^4]$

$f'(x)=6x(5-x)^4+3x^2[4(5-x)^3*(-1)]$

$f'(x)=6x(5-x)^4-12x^2(5-x)^3$

Now, we can factor $6x(5-x)^3$

$f'(x)=6x(5-x)^3[(5-x)-2x]$

exc...

3. Originally Posted by ecMathGeek
That solution is not possible. I went ahead and did some of the work (before noticing that the solution is not possible) so I'll post it anyways. However, the original function was a 6th degree polynomial (if you have multiplied it all out), but the solution you posted is a 20th degree polynomial. That's clearly wrong.

$f(x)=3x^2(5-x)^4$

$f'(x)=\frac{d}{dx}[3x^2]*(5-x)^4+3x^2\frac{d}{dx}[(5-x)^4]$

$f'(x)=6x(5-x)^4+3x^2[4(5-x)^3*(-1)]$

$f'(x)=6x(5-x)^4-12x^4(5-x)^3$

Now, we can factor $6x(5-x)^3$

$f'(x)=6x(5-x)^3[(5-x)-2x^3]$

exc...
the $12x^4$ should be $12x^2$ but that doesn't detract from what you are saying.

when we take the derivative of a polynomial, we cut the power down by 1