I currently in my first semester of Single Variable Calculus. I did reasonably well in Algebra (A), Trigonometry (B+), and Pre-Calculus (B+). However, I'm having some difficulty learning to differentiate more complex functions.

I remember formulas reasonably well, so I remember the various rules of differentiation: the power rule, the rule for finding derivatives of a sum/difference or constant multiple of a function, as well as the product, and quotient rules. I'm having more difficulty with the chain rule, though - I understand the basic formula:

$$ \frac{df}{dx} = \frac{df}{du}\frac{du}{dx} $$

But in practice, such as when trying to differentiate a complex function (like the one below), which involves using some or all of the differentiation rules in combination, I get bogged down in applying the rules correctly: i.e. in breaking the complex function into its constituent parts and applying the appropriate differentiation rules to them, in the context of the chain rule.

$$ f(x) = \frac{3x^7+x^4\sqrt{2x^5+15x^\frac{4}{3}-23x+9}}{6x^2-4} $$

With problems like this one (especially any differentiation problem that involves use of the quotient rule, and especially any problem requiring the use of a combination of the product, quotient, and/or chain rules), I have a lot of trouble keeping all the small, constituent functions straight and then combining the values of their derivatives correctly to arrive at the correct answer.

Surely, I can't be the first person to have this trouble. So, my question is: is there any intuition, technique, or approach, aside from "practice, practice, practice" (which I am doing) that might help me here?

I've tried using the Wolfram Alpha Calculus course assistant app to show me the differentiation steps when I get stuck, but it can be a bit hard to follow, for me. Sometimes it seems to use notation that is different from the notation my instructor uses in class.

  • 1
    $\begingroup$ I absolutely understand your trouble. I don't have a great answer at the moment, but I never bothered to remember the quotient rule, and write quotients as $f(x)g(x)^{-1}$, whose derivative becomes $f'(x)[g(x)^{-1}] - f(x)[g'(x)g(x)^{-2}]$. I know it's not much better, but it helped me a lot. In addition, use brackets, lots of brackets! Take things one step at a time, as agonizing as it is. $\endgroup$ – pjs36 Feb 18 '15 at 0:52
  • $\begingroup$ Thanks! Yep, I've tried that approach, also (using the product rule and writing quotients as the product of two fractions). It definitely works. I'm having a lot of trouble applying the chain rule especially, and keeping track of all the component functions in big, ugly functions like that one I posted. I edited my post to say I'm having trouble differentiating problems that require combinations of the quotient, product, and/or chain rules. Thanks for your reply. $\endgroup$ – tommytwoeyes Feb 25 '15 at 6:28

One technique that my students sometimes (claim to) find helpful in keeping all the small sub-functions straight is, simply, to name them. So write $f(x)=g(x)/h(x)$, where $g(x)=3x^7+x^4\sqrt{2x^5+15x^{4/3}-23x+9},h(x)=6x^2-4$. Then $f'=(hg'-gh')/h^2$, and we can plug in with $h'=12x$, $$f'=\frac{(6x^2-4)g'-(3x^7+x^4\sqrt{2x^5+15x^{4/3}-23x+9})12x}{(6x^2-4)^2}$$ Now $g$ is complicated enough that we should break it down another step. So write $g=3x^7+x^4 k\circ \ell$, where $k(x)=\sqrt{x}$ and $\ell(x)=2x^5+15x^{4/3}-23x+9$. Then $g'=21x^6+4x^3k\circ \ell+x^4(k\circ \ell)'=21x^6+4x^3k\circ\ell+x^4(k'\circ \ell)\ell'$. Now $k'=1/2x^{-1/2}$ and $\ell'=10x^4+20x^{1/3}-23$, so we have $$g'(x)=21x^6+4x^3\sqrt{2x^5+15x^{4/3}-23x+9}+x^4(1/2(2x^5+15x^{4/3}-23x+9)^{-1/2}))(10x^4+20x^{1/3}-23)$$ Now just copy-paste that in to your expression for $f'$ and we're done! The point of this technique is to break everything into bite-size steps which you can combine into the complete derivative completely mechanically: you have expressions for little functions and derivatives like $\ell'$ in terms of $x$, and expressions for bigger derivatives in terms of little functions and derivatives, so you just have to write some parentheses around the former and copy them into the latter. Hope this helps!

  • $\begingroup$ Thanks, Kevin. That's a great idea. I actually did try that when I attempted to solve the problem, but I didn't do quite as well as you did. I think you've presented the solution to my problem, and that I just need to practice. A lot. $\endgroup$ – tommytwoeyes Mar 2 '15 at 5:09

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