So I'm fairly new to this and I just wanted to check my understanding of the chain rule.

Suppose $f(x) = (ax + b)^n$ and we want to find $f'(x)$. We first work out the derivative of the first function, then multiply it to the derivative of the second function giving:

$$n(ax +b)^{n-1}\cdot(ax+b)'$$

To calculate the derivative of the second function, we use the sum rule, but this is where I get a little stuck. I know I have to find both $ax'$ and $b'$, so here's how I think I should do it...please tell me if I'm crazy and wrong:

As $b$ is a constant it follows that $b' = 0$. My textbook hasn't gone into this yet, so it isn't clear why. Anyway, moving on. This leaves us with:

$$ax' = a$$

What I REALLY don't understand is why we don't use the product rule for $ax$. Any clues?

  • $\begingroup$ That is because a is a constant so even if we consider the product rule, we have $a'\times x+ a\times x' =a$ as a is a constant. $\endgroup$ – user371838 Dec 2 '16 at 12:55
  • $\begingroup$ It may help if you let $f(x)=g^n(x)$ and $g(x)=ax+b$. $\endgroup$ – MathArt Dec 2 '16 at 12:59
  • $\begingroup$ Just an aside: $f(x) = b$ (where $b$ is a constant) implies $f'(x) = 0$ because, using the definition of the derivative, we have (note that $f(x) = b \implies f(x+h) = b$) \begin{align} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{b - b}{h} \\ &= \lim_{h \to 0} \frac{0}{h} \\ &= 0 \end{align} $\endgroup$ – mattos Dec 2 '16 at 13:32

In your first displayed equation, you have a $+$ which should be a $\cdot$.

You can use the product rule for $ax$. You get $a'x + ax'$. Then since $a$ is constant, $a'=0$, leaving you with $ax'$.

We math people are nothing if not consistent.

| cite | improve this answer | |
  • $\begingroup$ Well you have problem then mate, because it has been proven that you can't prove nor disprove that math is consistent. Edit: let me be more precise, the proof was about arithmetics (: $\endgroup$ – Itay.V Dec 2 '16 at 13:06
  • $\begingroup$ @Itay.V I think you have a fundamental misunderstanding of Goedel's theorem. We can be consistent as long as we're not complete. $\endgroup$ – B. Goddard Dec 2 '16 at 13:43

So the question seems to be how to take the derivatives of constants. There's no general answer. Of what we really take derivatives are not constants, but actions (aka as functions) instead. So a number $a$ is not a function itself, but it gives rise to several a actions.

For example, adding $a$ to a given function $f$ results in adding zero to the derivative: $ (f+a)'=f'+0$. The action "${}+a$" changes to the action "${}+0$".

Multiplying $a$ with $f$ yields $(a\cdot f)'=a\cdot f'$, that is, the action "$a\cdot{}$" has magic hat on regarding taking derivatives, the action "$a\cdot{}$" stays unchanged after taking the derivative.

Since $x'=1$ we'll find $(a\cdot x+b)'=a\cdot1+0=a$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.