3.13: Multirule Derivatives
Okay, let’s talk about \(\frac{d}{dx} \ e^{x^2 + x} \sin(x)\). If you’re thinking this looks like a product rule, but it also looks like a chain rule, you’re right. To compute this derivative, we need to do the chain rule and the product rule. This is because it is a multirule problem. Let’s do this example
The way I like to break this down is to consider a little rule and a big rule. In this case, the little rule is the chain rule problem \(\frac{d}{dx} \ e^{x^2 + x}\). If we do this problem, we see that \(f = e^x\), \(f' = e^x\), \(g = x^2 + x\) and \(g' = 2x + 1\). So we have
\[\begin{equation*} \frac{d}{dx} \ e^{x^2 + x} = {\color{red} e^{x^2 + x} (2x + 1)}. \end{equation*}\]
Now we are ready to do the big rule, which is the product rule. At this point we go back to the original problem \(\frac{d}{dx} \ e^{x^2 + x} \sin(x)\). For this product rule, we see \(f = e^{x^2 + x}\), \(g = \sin(x)\), \(g' = \cos(x)\). What is \({\color{red} f'}\)? Why, that’s what we just computed in the equation above! So \({\color{red} f' = e^{x^2 + x}(2x+1)}\). Putting this all together with the product rule \(f g' + g f'\), we have
\[\begin{align*} \frac{d}{dx} \ e^{x^2 + x} \sin(x) & = f g' + g {\color{red} f'} \\ & = \boxed{e^{x^2 + x} \cos(x) + \sin(x) {\color{red} e^{x^2 + x} (2x + 1)}} . \end{align*}\]
little chain rule: \(\frac{d}{dx} \sin(x^2 + x)\)
\(\begin{array}{ll} f = \sin(x) & g = x^2 + x \\ f' = \cos(x) & g' = 2x + 1 \end{array}\)
Result: \({\color{red} \cos(x^2 + x) \cdot (2x+1)}\)
Big quotient rule (aka the whole problem):\(\frac{d}{dx} \frac{x}{\sin(x^2 + x)}\)
\(\begin{array}{ll} f = x & g = \sin(x^2 + x) \\ f' = 1 & g' = {\color{red} \cos(x^2 + x) \cdot (2x+1)} \end{array}\)
Result: \(\boxed{\frac{\sin(x^2 + x) \cdot 1 - x \cos(x^2 + x) \cdot (2x+1)}{(\sin(x^2+x))^2}}\)