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I am wary of getting in over my head, after initial searching it is apparent to me that I don't know how to properly phrase this question for the answer I want. I am only working at high school level but in class we learnt differentiation and how when

$y = ax^n$,

$ \frac{dy}{dx} = anx^{(n-1)}$.

I was wondering if this applies to when

$y = 1$,

$\frac{dy}{dx} = 1\times1^0 = 1$.

and also when

$y = 1^2$,

$\frac{dy}{dx} = 2\times1^1 = 2$.

Obviously the gradient should be 0, and when calculated in terms of x it makes sense

$y = x^0$,

$\frac{dy}{dx} = 0\times \frac{1}{x} = 0$.

I was wondering how this should be approached, since to me it implies that you CANNOT have a line without a $y$ AND $x$ term, yet $y=1$ CAN be drawn.

Is y=1 a different thing to $y=x^0$? Does $y=1$ even exist in 2D space? Or do you just have to simplify your equation before differentiating it?

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    Please use MathJax/LaTeX formatting so we can read your question and help you more quickly! – 2017-02-17
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    We cannot differentiate numbers, but only functions. Thus $y=1$ is the constant function that maps every value of $x$ to $1$. Every constant function $y=c$ has as derivative the function $y'=0$. – 2017-02-17
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    Why are you talking about gradients and 2d space? All the functions you're discussing are single-variable functions. – 2017-02-17
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    any constant $c$ can be written as $cx^{0}$ and upon differentiating wrt $x$ gives $0$ as you point out. The process of finding a derivative concerns small rates of change of a function of real (or complex) variables. – 2017-02-17
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    The formula is valid for all $n$. – 2017-02-17

4 Answers 4

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The other answers are correct, but they miss the bigger error you made: you write

since $y=1^2$, we have ${dy\over dx}=2\cdot 1^{2-1}$.

But this is misapplying the power rule! Remember that this rule says $${d\over dx}(x^n)=nx^{n-1},$$ but - in the highlighted step above - you've conflated $x$ and $1$! (Note that the issue with $n=0$, while real, doesn't even enter here - the error is more basic than that.) If you apply the power rule "correctly" - that is, just matching up symbols in the obvious way - you get

since $y=1^2$, we have ${dy\over d1}=2\cdot 1^{2-1}$.

But of course "${dy\over d1}$" doesn't make sense - that's asking, "How does $y$ change as $1$ changes?" But $1$ can't change.

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$$y=x^{0}$$ is only defined if $$x\ne 0$$ but $$y=1$$ is defined for all real $x$ if $$y=1$$ then we get the first derivative as zero, since $1$ is constant

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    While correct, this isn't the main error - they've also conflated $1$ and $x$ in the line "$y = 1^2$, $\frac{dy}{dx} = 2\times1^1 = 2$." – 2017-02-17
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In calculus you differentiate functions, not formulas. Much of your confusion comes from thinking about variables $x$ and $y$ instead of functions. But since that's how you're learning I will try to use that vocabulary.

The function you have in mind when you write $y=1$ is the function whose value is always $1$, independent of the value of the independent variable which you are implicitly thinking of as "$x$". The graph of that function is a horizontal line of height $1$. Its slope at every point is $0$, so the derivative of the function is $dy/dx=0$.

Some functions that you encounter have formulas. For example, the formula $$ y= x $$ has graph a straight line sloping up from the origin while $$ y= x^2 $$ describes a parabola.

Early in the study of calculus you show that when a function is given by a formula $$ y = x^n $$ for a positive integer $n$ then its derivative is given by the formula $$ \frac{dy}{dx} = n x^{n-1}. $$

Later in your study you will show that formula works for all values of $n$ (not just integers) except for $n=0$. In that case you know the answer right away because the intent it to describe a constant function, whose graph is horizontal. The formula just happens to give the the right answer except when $x=0$, but you shouldn't use the formula. That's the long answer to your good question.

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    While correct, this isn't the main error - they've also conflated $1$ and $x$ in the line "$y = 1^2$, $\frac{dy}{dx} = 2\times1^1 = 2$." – 2017-02-17
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    @NoahSchweber You address that in your answer (+1). Perhaps it's the "main error". Thinking about functions rather than formulas would prevent it too. – 2017-02-17
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It's perfectly logical to take the derivative of $1$ with respect to $x$, and you get the same answer as when you take the derivative of $x^0$ with respect to $x$.

The reason you're getting a different answer is that you've made a mistake when calculating the derivative of $1$. It looks like you're trying to apply this rule:

If $y = 1^n$, then $\frac{dy}{dx} = n\ 1^{n-1}$. (incorrect)

It is true that if $y = x^n$, then $\frac{dy}{dx} = n\ x^{n-1}$. But since $1$ isn't $x$, you can't apply the rule to $1$ like that.

One way of calculating the derivative correctly is by using the chain rule. Suppose we define $f(x) = 1$; $g(z) = z^n$; and $y = g(f(x)) = 1^n$. Then

$$\frac{dy}{dx} = g'(f(x)) \cdot f'(x) = n\ (f(x))^{n-1} \cdot 0 = 0.$$

To answer your little questions:

Is y=1 a different thing to y=x^0? Does y=1 even exist in 2d space? Or do you just have to simplify your equation before differentiating it?

The equations $y = 1$ and $y = x^0$ mean the same thing (except that $x^0$ may be undefined when $x = 0$, depending on who you ask). The equation $y = 1$ does, in fact, define a curve in 2D space; it's a line with $y$-intercept 1. It is not necessary to simplify your question before differentiating it; if you do the differentiation correctly, you'll get the same answer either way.