In the solved example above, rather than taking $a^2x^4$ together and differentiating $a^2 = 0$, we differentiated $x^4$ and took out $a^2$. Why? Couldn't we have differentiated $a^2$ and gotten the answer zero?
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$\begingroup$We have $\dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2 x^4\right) = a^2 \cdot 4x^3$. You can see that this is true by using the product rule as follows:
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2 x^4\right) = x^4 \dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2\right) + a^2\dfrac{\mathrm{d}}{\mathrm{d}x} \left(x^4\right)$$
Like you said $\dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2\right) = 0$, so the above simplifies to $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left(a^2 x^4\right) = a^2\dfrac{\mathrm{d}}{\mathrm{d}x} \left(x^4\right) = a^2 \cdot 4x^3$$
This is true in the general case, let $f(x) = a\cdot g(x)$ where $a$ is a constant and $f, g$ are differentiable functions. Then using the notation that $f'(x)$ is the derivative of $f$ with respect to x and the product rule, we have $$f'(x) = g(x)\dfrac{\mathrm{d}}{\mathrm{d}x}\left(a\right) + a \cdot g'(x)$$
Applying the rule that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(a\right) = 0$ simplifies the above to $$f'(x) = a\cdot g'(x)$$
As a side note, whilst it is true that $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(f(x) + g(x)\right) = \dfrac{\mathrm{d}}{\mathrm{d}x}\left(f(x)\right) + \dfrac{\mathrm{d}}{\mathrm{d}x}\left(g(x)\right) = f'(x) + g'(x)$$
which is called the linearity of the differential operator, we have $$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(f(x) \cdot g(x)\right) \neq \dfrac{\mathrm{d}}{\mathrm{d}x}\left(f(x)\right) \cdot \dfrac{\mathrm{d}}{\mathrm{d}x}\left(g(x)\right) $$
$\endgroup$ 6 $\begingroup$Yes, you can always take the constant out. You are right in stating that $$\frac{d}{dx}(a^{2})=0 $$ But be aware of the fact below.
$$ \frac{d}{dx}(a^{2}x^{4})\neq \frac{d}{dx}(a^{2}) \cdot \frac{d}{dx}x^{4} $$
If you want to do $\frac{d}{dx}(a^{2}x^{4})$, apply chain rule and you will obtain $$ \frac{d}{dx}(a^{2}x^{4})=a^{2}\frac{d}{dx}x^{4} $$ So whenever you have a constant, you can just take it out.
$\endgroup$ $\begingroup$Derivative of a anything multiplied by a constant is the derivative of that thing times that constant. You can jolly well take out the constant in the multiplicative context.
$$ \dfrac {d (b^2)}{dx} = b^2 (0) $$
is nonsense. Derivative of a constant is directly zero.
$\endgroup$ 3 $\begingroup$One thing about the differentiation rules is that they always work in any circumstance.
Yes, you can always pull out the constant while differentiating, but at the same time, since $f(x)=a$ is certainly a function, taking the derivative of $ax^{2}$ using the product rule had better still work - it's a product of functions and so the product rule must work. It does and must work, it's just that it's quicker to use the rule that constants can be pulled out of the differentiation operation, but these two methods will always agree, they are equivalent.
Similarly, if I asked you to differentiate $x^{2}$, you would probably quickly use the power rule and return $2x$, but what if we wrote this as $x \cdot x$? Since it's the same as $x^{2}$, and it's a product of functions, the product rule had better work here, and it does: $\frac{d}{dx}(x \cdot x) = x \cdot \frac{d}{dx} x + \frac{d}{dx}x \cdot x = x + x = 2x$.
Pretty much any function can have multiple rules applied to it to take the derivative. The derivative rules are complete mathematical theorems, they will never fail to work.
If you want to pull the constant out of differentiation, do so, if you want to use the product rule, do so; both are valid moves. Note that using the product rule on a constant function and arbitrary function product proves that pulling the constant out is always valid.
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