# 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.

• $$1^{a} = 1 \quad \text{for all } a \in \mathbb{R}.$$
– MSDG
Nov 26, 2018 at 21:49
• @Shaun a circle is not a function Nov 26, 2018 at 21:52
• You are taking derivatives of two different functions. it is no surprise you get two derivatives. BTW, the convention is for positive number $K$, $K^{1/n}$ always refers to the positive $n$-th root of $K$. Nov 26, 2018 at 22:10
• If you want to consider all roots, $1^{1/4}=1,i,-1,-i$, and not $\pm1$.
– user65203
Nov 26, 2018 at 22:58
• @Spitemaster: It is false. Please don't confuse students. $9^{1/2} = 3$ full stop, by definition of exponentiation as stated in Shubham's answer. Nov 28, 2018 at 6:00

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, $$y = \frac{ 4^{1/5} }{5} x^{1/5} - { \frac{1}{x ^ {3/4}} }$$.

• but $x^{1/4} = \pm1$, isn't it? shouldn't it be proper to include both? Sorry for not understanding Nov 26, 2018 at 22:00
• @Jaacko In that case it is not a function. Taking derivatives is something you can only do to functions. Nov 26, 2018 at 22:04
• Let us say $f(x)=1^{\frac{1}{4}}$. For $f$ to be a function, all $x$ must have unique images. If you were to say $f(x)=\pm1$, all points will get mapped to both $1$ and $-1$. Is this a function? Nov 26, 2018 at 22:06
• One last thing and this is going just a tad outside of the question, but shouldnt $y = \frac{dy}{dx} = \frac{ 4^{1/5} }{25} x^{-4/5} - \frac{3}{4} { x ^ {-7/4} }$ be a perfectly good answer as the derivative of the original function? Nov 26, 2018 at 22:07
• Absolutely. It is one of the answers. But you need to be careful here. Some operations, such as the square root, are conventionally defined as being positive valued so that they may be treated as functions. Then, it is not meaningful to interpret it as negative valued. Nov 26, 2018 at 22:08

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.

• This makes a lot more sense! Sorry I already gave the answer to someone else, but this is something I shall not forget Nov 26, 2018 at 22:14
• I definitely would not say it’s “laziness” or that it’s “misunderstanding.” The $\pm$ notion with roots is essential in, for example, the quadratic formula. Nov 27, 2018 at 0:52
• @ChaseRyanTaylor It's not "essential" in the quadratic formula. You could say the solutions to $ax^2+bx+c=0$ are $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$, so it's not "essential" to use $\pm$. Using $\pm$ there is more efficient for those who understand what it really means. But as you can see by the OP and many students, use of $\pm$ leads to conceptual misunderstandings. I see many students who think that $\sqrt{4}=\pm2$ for example. Nov 27, 2018 at 3:19
• @JaackoTorus you are allowed to change your accepted answer if you want to.
– SQB
Nov 28, 2018 at 8:45
• The accepted answer is good, and seems to me to more directly address the OP question. My answer here notices an issue with the OP's reasoning and speaks to that issue, but imho is less directly answering the question. So I support the accepted answer, even though I could nitpick some of its details (which surely can be done with my answer too). Nov 28, 2018 at 21:53

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).

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 $$\forall r\in \mathbb R^ * \quad \frac{\text{d}x^r}{\text{d}x}=rx^{r-1}$$

• But $1^{1/n}=1$, never $-1$ ! Any real power of a real number is positive, except for $0^0$ which is undefined. Nov 26, 2018 at 22:55
• "Any real power of a real number is positive" is wrong. $(-1)^1=-1$. Nov 27, 2018 at 10:21
• Damn ! Too late to edit, meant positive number ! Nov 27, 2018 at 13:58
• $(-1)^n=\pm1$ is a statement about the value varying when $n$ varies. For a fixed $n$, $(-1)^n$ is fixed. Nov 27, 2018 at 23:16

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.