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Find the derivative:

$$\frac{d}{dx}\left[(x^3-x^2+5)(x^4-3x^2+7x)\right]$$

The answer is:

$$(3x^2 - 2x + 5)(x^4-3x^2 + 7x) + (x^3-x^2 +5x)(4x^3-6x+7)$$

Why is the function part of the answer?

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    Not sure what is your question. The derivative maps a function to a different but related function. your 7 th degree polynomial has mapped to a 6th degree polynomial. – 2017-02-03
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    It is not. I do not see $x^7 - x^6 + 2 x^5 + 10 x^4 - 22 x^3 + 35 x^2$ anywhere. – 2017-02-03

2 Answers 2

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The following is the template for taking the derivative of a product of functions:

$$\frac{d}{dx}\big (f(x)\cdot g(x)\big) = f'(x)\cdot g(x) + f(x)\cdot g'(x)$$

In your case, we have:

$$\frac d{dx}\left[\underbrace{(x^3 -x^2 + 5x)}_{\large f(x)}\underbrace{(x^4-3x^2+7x)}_{\large g(x)}\right]$$

by the product rule for derivatives,

$$= \underbrace{(3x^2 - 2x + 5)}_{\large f'(x)}\underbrace{(x^4-3x^2 + 7x)}_{\large g(x)}\;\; + \;\;\underbrace{(x^3 - x^2 + 5x)}_{\large f(x)}\underbrace{(4x^3 - 6x+7)}_{\large g'(x)}$$

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Hint.

Treat your function as the product of two functions $f$ and $g$. Hence

$$\frac{\text{d}}{\text{d}x} \Big(f(x)\cdot g(x)\Big) = f'(x)g(x) + f(x)g'(x)$$

I think it's straightforward to understand this.