Derivative of $f(x) = (x^2 +1)^3 (2x+5)^2$ I have function
$$f(x) = (x^2 +1)^3 (2x+5)^2$$
I need to find the derivative.
I believe that I need to use the product rule and chain rule.
Here's what I did.
$$f'(x) = (2x+5)^2[3(x^2+1)^2(2x)] + (x^2+1)^3[2(2x+5)2]
\\ = (2x+5)^2(6x(x^2+1)^2)+4(2x+5)(x^2+1)^3
\\ = 6x(2x+5)^2(x^2+1)^2+4(2x+5)(x^2+1)^3$$ 
That is what I have.
But the problem is, the answer book says that the answer is,
$$2(x^2+1)^2(2x+5)(8x^2+15x+2)$$
Since the answer book has been wrong few times, so I checked the answer with a calculator.
The calculator showed the same answer.
Then I used another calculator (which showed some steps), showed my answer.
The calculator showed, 
Which answer is correct?? Or are they just equivalent answers??
If I did wrong, what was the problem??
Can it be factorized? How?
Thank you
 A: Factor out common terms:
$$ 6x(2x+5)^2(x^2+1)^2+4(2x+5)(x^2+1)^3 = 2(2x+5)(x^2+1)^2(3x(2x+5) + 2(x^2 + 1))$$
Simplify to get $$2(2x+5)(x^2+1)^2(6x^2+ 15x + 2x^2 + 2)= 2(2x+5)(x^2+1)^2(8x^2+ 15x + 2)$$
The text book is correct, AND your work was fine (take comfort in that). It was just a matter of simplifying your work through factorization. 
A: This is typical case where logarithmic differentiation makes life easy. $$f(x) = (x^2 +1)^3 \times(2x+5)^2$$ $$\log\big(f(x)\big)=\log(x^2+1)^3+\log(2x+5)^2=3\log(x^2+1)+2\log(2x+5)$$ Now, differentiate both sides $$\frac{f'(x)}{f(x)}=3\times \frac{2x}{x^2+1}+2\times\frac{2}{2x+5}$$ Now, reduce to same denominator $$\frac{f'(x)}{f(x)}= \frac{6x(2x+5)+4(x^2+1)}{(x^2+1)(2x+5)}=\frac{16 x^2+30 x+4}{(x^2+1)(2x+5)}=2\frac{8 x^2+15 x+2}{(x^2+1)(2x+5)}$$
Multiply now both sides by $f(x)$ and simplify to get $$f'=2(8 x^2+15 x+2)(x^2+1)^2(2x+5)$$ which is the answer from the book.
Do you find this way easier ? 
A: Factor out common terms:
$6x(2x+5)^2(x^2+1)^2+4(2x+5)(x^2+1)^3=2(2x+5)(x^2+1)^2[3x(2x+5)+2(x^2+1)]=$
$$2(2x+5)(x^2+1)^2[6x^2+15x+2x^2+2]=$$
$$2(2x+5)(x^2+1)^2(8x^2+15x+2)$$
both the textbook and you are correct.
