Can a function have two derivatives? I am a senior in high school so I know I am simply misunderstanding something but I don't know what, please have patience.
I was tasked to find the derivative for the following function:
$$ y = \frac{ (4x)^{1/5} }{5} + { \left( \frac{1}{x^3} \right) } ^ {1/4} $$
Simplifying:
$$ y = \frac{ 4^{1/5} }{5} x^{1/5} + { \frac{1 ^ {1/4}}{x ^ {3/4}} } $$
$$ y = \frac{ 4^{1/5} }{5} x^{1/5} + { \frac{\pm 1}{x ^ {3/4}} } $$
Because $ 1 ^ {1/n} = \pm 1 $, given $n$ is even
$$ y = \frac{ 4^{1/5} }{5} x^{1/5} \pm { x ^ {-3/4} } $$
Taking the derivative using power rule:
$$ \frac{dy}{dx} = \frac{ 4^{1/5} }{25} x^{-4/5} \pm \frac{-3}{4} { x ^ {-7/4} } $$
which is the same as
$$ \frac{dy}{dx} = \frac{ 4^{1/5} }{25} x^{-4/5} \pm \frac{3}{4} { x ^ {-7/4} } $$
And that is the part that I find difficult to understand. I know that I should be adding the second term(I graphed it multiple times to make sure), but I cannot catch my error and my teacher did't want to discuss it.
So I know I am doing something wrong because one function cannot have more than one derivative.
 A: You are absolutely right to question this. This is what a good mathematician does.
The exercise does not define the function properly. For a proper definition it should have mentioned the set the function is defined on and the image set. Something like: $f$ defined as a mapping $f:\mathbb{R_+} \to \mathbb{R_+}$ with $f(x) =\ldots$. 
This would have ruled out the $-1$ interpretation (letting $x$ go to zero would then contradict the definition of the function). 
A: The $(\cdot)^{\frac{1}{4}}$ operation has to be understood as a function. A function can only have one image for any argument. Depending upon how you interpret the fourth root, the image could be positive or negative. But once you set how you interpret your function (positive or negative valued), you have to stick with that interpretation throughout. 
When you write $y = \frac{ 4^{1/5} }{5} x^{1/5} \pm { \frac{1}{x ^ {3/4}} }$, you are working with both interpretations simultaneously. In other words, when you differentiate, you don't get two derivatives for one function, rather two derivatives corresponding to two different functions, one $y = \frac{ 4^{1/5} }{5} x^{1/5} + { \frac{1}{x ^ {3/4}} }$, and the other, $y = \frac{ 4^{1/5} }{5} x^{1/5} - { \frac{1}{x ^ {3/4}} }$.
A: You are confused about what $y^{1/4}$ actually means.
Suppose that $x^4=1$. We could raise both sides to the $1/4$ power: $$\left(x^4\right)^{1/4}=1^{1/4}$$
The right side is unambiguously $1$. It is not $\pm1$. But read on. $$\left(x^4\right)^{1/4}=1$$ The left side does not simplify to $x$ unless you somehow know ahead of time that $x$ is positive. Otherwise, all you can say is the left side simplifies to $\lvert x\rvert$. So you have $$\lvert x\rvert = 1$$ That implies that "either $x=1$ or $x=-1$". Out of laziness (or a minor efficiency boost) people write $x=\pm1$.
Now we started with $x^4=1$ and ended with $x=\pm1$. And because of this and applying the $1/4$ power in the middle of that process, you have inferred that $1^{1/4}=\pm1$. But that is a misunderstanding of the process in its entirety. $1^{1/4}$ is unambiguously equal to $1$ when working with arithmetic and real numbers.
A: Your mistake is in the statement $1^{1/n}=\pm 1$.
First of all this is wrong, two ways to see this are :
$1^{1/n}=\sqrt[n]{1}=1$ or $1^{1/n}=e^{-n\ln{1}}=e^0=1$.
You may be confusing this with $(-1)^n=\pm 1$.
Secondly, even if $1^{1/n}$ varied with $n$, here you have a given value for $n$ which is $n=4$ which would determine whether the answer is $1$ or $-1$.
Although the equation $a^4=1$ does indeed have two real solutions, $1^{1/4}$ denotes the positive one, and that's what is implied in the formula for differentiating power functions that you used, known as $\forall r\in \mathbb R^ * \quad \frac{\text{d}x^r}{\text{d}x}=rx^{r-1}$
A: When you have a root, such as $\sqrt x$ or $x^{\frac14}$, that's generally taken to be the what's called the "principal root". For positive numbers, the principal root is positive. So $1^{\frac14}$ is just 1. Where you need to keep track of the $\pm$ is when you apply a root to both sides: of an equation if $x^2=1$, then $x=\pm1$. This is because the rule $(x^a)^{\frac1a}=x$ can fail if $x$ is negative. 
