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From my understanding, we can't use the power rule when $x$ is being modified. So how can we use power rule on derivative of $(x+1)^2$? I know that we could expand it and find the derivative of each part (and add these together), but I receive the same answer when bring down the two and multiply through. Is this just a coincidence?

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  • $\begingroup$ Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ Feb 20, 2018 at 22:00
  • $\begingroup$ Chain rule..... $\endgroup$ Feb 20, 2018 at 22:02
  • $\begingroup$ No... if $\dfrac{d}{dx}(x+1)^2$ wasn't equal to $\dfrac{d}{dx}(x^2+2x+1)$ then you'd break math. $\endgroup$
    – Andrew Li
    Feb 20, 2018 at 22:03
  • $\begingroup$ What you did works only when the derivative of the enclosed function is equal to $1$. That won't work the same if you only use the power rule to differentiate $(5x+3)^2$, or $(x^3+5)$ $\endgroup$
    – amWhy
    Feb 20, 2018 at 22:11
  • $\begingroup$ @Dude156 You are using power and chain rules. When you are using power rule, you are not touching base $(x+1)$ but when differenting $(x+1)$ in chain rule as a second necessary operation you can change it. $\endgroup$
    – Narasimham
    Feb 20, 2018 at 23:25

3 Answers 3

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Chain rule, let $y=x+1$,then \begin{align}\frac{d}{dx}(x+1)^2&=\frac{d}{dx}y^2\\&=\frac{d}{dy}y^2\frac{dy}{dx}\\&=2y \frac{dy}{dx}\\&=2y\\&=2(x+1) \end{align}

Of course, eventually, we no longer write down $y$ explicitly after sufficient practice.

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  • $\begingroup$ So why can't we use a similar method when finding the derivative of sec x? $\endgroup$
    – Dude156
    Feb 20, 2018 at 22:10
  • $\begingroup$ hmmm... what do you mean by similar method when finding $\frac{d}{dx} \sec x$? $\endgroup$ Feb 20, 2018 at 22:15
  • $\begingroup$ Ok so say we write $sec(x)$ as $1/cos(x)$ now we replace cos(x)=y. Now we take the derivative of $y^(-1)$. This gives us $-y^(-2)$. Thus substituting back cos(x) we get $-sec(x)^2$. This is incorrect though, as the correct answer should be $tan(x)*sec(x)$ $\endgroup$
    – Dude156
    Feb 20, 2018 at 22:18
  • $\begingroup$ You also need to take the derivative of $\cos x$, so the derivative is $$\frac{-\sin x}{-\cos(x)^2} = \tan x \cdot \sec x$$ So $$(\sec x)' = (\frac 1{\cos x})' = (\cos x)'\cdot (\cos^{-1} x)' = \tan x \cdot \sec x$$ $\endgroup$
    – amWhy
    Feb 20, 2018 at 22:31
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The answer is that you can use the power rule sometimes if the $x$ is being modified. To be as thorough as possible, I'll assume you haven't learned about Chain Rule yet.

Remember that when you find the derivative of, say, $x^3$, you're finding the rate of change of that function. Now take something like $(3x^2+1)^4$; it makes sense that the rate of change of this should depend not only on that $x^4$ nature of the function, but also on the rate of change of that part inside the parentheses. There is a more specific way to say this, called Chain Rule.

The Chain Rule says that when you have a composition of two functions, that is, $f(g(x))$, then the derivative is given by $$\frac{d}{dx}f(g(x))=g'(x)f'(g(x))\text{, or }\frac{dg}{dx}*\frac{df}{dg}$$ This is more often remembered as "the inside times the outside". The proof for this is a bit involved, but Wikipedia has 3 different versions if you're interested.

For example, let's take the derivative of that function from before, $(3x^2+1)^4$. Chain rule says we take the derivative of the inside, $6x$, and multiply it by the derivative of the outside (ignoring the inside), which is $4(3x^2+1)^3$. This gives us our answer: $$\frac{d}{dx}(3x^2+1)^4=6x*4(3x^2+1)^3=24x(3x^2+1)^3$$

Now we can return to your observation. Let's apply chain rule to your function, $(x+1)^2$: $$\frac{d}{dx}(x+1)^2=1*2(x+1)^1=2x+2$$ Now you can see why the Power Rule appears to work on this: when we use Chain Rule, the derivative of the inside is just 1, so it doesn't affect the result. Instinctively, this makes sense, since the whole point of Chain Rule is to account for how much the inside changes, and $x+1$ changes just as fast as $x$.

Notice that there isn't anything special about $x+1$. We could have used anything which has a derivative of $1$, like $x+3$, or even $x+999,999$. But be careful; as soon as the derivative inside is no longer $1$, like with $2x+1$ or $\frac{x}{2}+3$, Chain Rule is no longer trivial.

EDIT: As an addendum, I'll demonstrate chain rule using the example from the comments on another answer, $\sec(x)$. $$\frac{d}{dx}\sec(x)=\frac{d}{dx}(\cos(x))^{-1}=-\sin(x)*-1(\cos(x))^{-2}=\frac{\sin(x)}{\cos^2(x)}=\tan(x)\sec(x)$$

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Claim $$ \frac{d}{dx}(x+c)^n=n(x+c)^{n-1} $$ for all $c\in\mathbb{R}$. Indeed by the binomial theorem $$ (x+c+h)^n=(x+c)^n+n(x+c)^{n-1}h+\binom{n}{2}(x+c)^{n-2}h^2+\dotsb+h^n $$ so $$ \begin{align} \lim_{h\to0}\frac{(x+c+h)^n-(x+c)^n}{h}&=\lim_{h\to0}\,[n(x+c)^{n-1}+\binom{n}{2}(x+c)^{n-2}h+\dotsb+h^{n-1}]\\ &=n(x+c)^{n-1} \end{align} $$

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