Derivative rules I have some problems when I have to derive: I don't know where I have to begin and where I have to stop. For example, if I have $$f(x)=(x^2 +1)(x^2 +3)$$ I know that I have to use the product rule so I get $$f'(x)=(x^2 +1)'(x^2 +3)+(x^2 +1)(x^2 +3)'$$ and the resolution is $$f'(x)=4x^3 +8x$$. But why can't I derive the stuff inside the brackets, like $$f(x)'=(2x)(2x)$$ and then $$f'(x)=4x^2$$
And I always have that problem, I don't know what rule I should use first.
Thank you.
 A: You can't just 

derive the stuff inside the brackets

because that's not how derivatives work. The rate at which a product $AB$ changes when $A$ and $B$ change is not simply the product of the rates of change of $A$ and $B$. The correct way to calculate that is with the product rule. 
To think intuitively about the product rule, imagine that you make $\$100$/hour and you work for $10$ hours. To calculate the change in your earnings if you increase your rate by $\$1$/hour and your hours by $1$ hour you don't make just an extra $\$1$, you collect $\$101$ for all $11$ hours. That's an extra $\$111$.
Calculus is more than remembering just what "rule" to apply.
A: Since$$f(x)=(x^2+1)(x^2+3)=x^4+4x^2+3,$$you know that$$f'(x)=4x^3+8x,$$which is not $4x^2$. So, don't use the rule $(g\times h)'=g'\times h'$, because that's no rule at all. As you can see from this example it just doesn't work.
A: Write $g(x) = x^2+1$ and $h(x) = x^3+3$.
Then your $f$ is
$$ f(x) = g(x)h(x)$$
So the product rule says 
$$
f'(x) = g'(x)h(x) + g(x)h'(x).
$$
The second thing you wrote would be equivalent to 
$$
f'(x) = g'(x)h'(x)
$$
and this is just not how the derivative works.
A: Written as in your question $f(x)$ is a product of two functions. In that case you must apply the product rule: $(u(x)v(x))'=u'(x)v(x)+u(x)v'(x)$.
You could also work out the brackets, leading to:$$f(x)=x^4+4x^2+3$$
Written like that $f$ can be recognized as a sum of functions. Then it it is time to use the rule $(u(x)+v(x))'=u'(x)+v'(x)$
So a good thing to ask yourself is: "are we dealing with a product here of with a summation?".
Further there is no rule at all that states that $(u(x)v(x))'=u'(x)v'(x)$. So if that "rule" is part of your luggage then you must throw it away immediately!
A: You can use the rule $(uv)' = u'v+uv'$.
If $f(x)=4x^2$ then:
$f'(x) = (2x)'(2x)+(2x)(2x)' = 2(2x)+(2x)2 = 8x$
which is what you want.
A: Apply the product rule or try to simplify the expression first.
$$(uv)' = uv' + vu'$$
[Assuming that you mean, $f(x) = 4x^2 $ and not $f'(x) = 4x^2$, as you stated you want to find the derivative].
$f(x) = (2x)(2x)$
The derivative of $x$ is $1$ and $2$ is a constant.
So, $f'(x) = 2.1(2x) + 2x.2.1 =  4x + 4x = 8x$
Or by simplification,
$f(x) = (2x)(2x) = 4x^2$
$f'(x) = 4.2x = 8x$
A: You have two options to take the derivative of $$f(x)=(x^2+1)(x^2+3)=x^4+4x^2+3$$
1) Multiply first and take the derivative.
This method gives you $$f'(x) = 4x^3 + 8x$$
2) Apply product rule and simplify as you did and you get the same answer as the first option.
Note that  derivative of the product is NOT product of derivatives.  
There is only one product rule and it is:  derivative of product is derivative of the first function multiplied by original of the second function plus derivative of the second function multiplied by original of the first function.
A: You apply the differentiation rule in the same order as when you evaluate the expression.
$$(x^2+1)(x^2+3)$$ is a product.
The left factor, $x^2+1$ is a sum. The left term of this sum is a power, $x^2$, and right term a constant, $1$.
The left factor, $x^2+3$ is a sum. The left term of this sum is a power, $x^2$ and right term a constant, $3$.

To differentiate the expression, you first differentiate the product, and the rule will ask you to provide the derivatives of the factors.
To differentiate the left factor, you apply the rule for a sum, which will ask you to differentiate the terms.
And so on.

In fact you follow the priority order between the operators and parenthesis.
A: For any non-constant function $g(x)$. When we multiply $g(x)$ with a constant (the constant becomes coefficient), we are multiplying the slopes everywhere of the line by the coefficient. For example, $y=2x$ has the slope of $y=x$ multiplied by $2$.
However, it will not make any sense to take the derivative of a coefficient that intensifies the rate of change. Even when the coefficient changes with respect to $x$ (The multiplier will no longer be constant or coefficient, but multiplication still occurs), it still does not make sense.
There is another thing you can consider. Suppose you don't know the Product Rule. Instead of taking the derivative of the polynomials within brackets, why don't you try to expand the product?
$\frac{d}{dx}[(x^2+1)(x^2+3)]\\=\frac{d}{dx}(x^4+3x^2+x^2+3)\\=\frac{d}{dx}(x^4+4x^2+3)\\=\frac{d}{dx}x^4+\frac{d}{dx}4x^2+\frac{d}{dx}3\\=4x^3+8x$
Or evaluate the derivative using First Principle, though it will be tough.

I don't know what rule I should use first.

This is an invalid problem, because $[f(x)g(x)]'\neq f'(x)g'(x)$.
