How do I differentiate a product $a(x) b(x) c(x)$ of three functions?

Question

If I have a function $$f(x)=a(x)\cdot b(x)\cdot c(x),$$ where I cannot simplify any further, how do I differentiate it?

For example, how would I differentiate $$f(x)=(x^2+5x+6)\cdot(e^{2x})\cdot(\cos(\pi+x))\ ?$$

Answer 1

Hint Given the three (differentiable) functions $a, b, c$, we can suggestively rewrite their product as $$abc = (ab)c .$$ Then, applying the product rule to the factor functions $ab$ and $c$ gives $$(abc)' = [(ab) c]' = (ab)' c + ab c' ,$$ which reduces the computation to determining the derivative $(ab)'$ in terms of $a$ and $b$ and their derivatives, but this is exactly the content of the usual product rule that you already know.

Answer 2

HINT: we have $$f'(x)=a'(x)b(x)c(x)+a(x)b'(x)c(x)+a(x)b(x)c'(x)$$ for your example we get $$f'(x)=(2x+5)\exp(2x)\cos(\pi+x)+(x^2+5x+6)\exp(2x)2\cos(\pi+x)+(x^2+5x+6)\exp(2x)\cdot (-\sin(x+\pi))$$

Answer 3

A classic trick is by using implicit differentiation:

Given that

$$\log f(x) = \log(x^2+5x+6)+2x+\log(\cos(x+\pi))$$

we differentiate wrt $x$ to get

$$\frac{f'(x)}{f(x)} = \frac{2x+5}{x^2+5x+6}+2+\frac{-\sin(x+\pi)}{\cos(x+\pi)}$$

That is,

$$f'(x) = f(x)\left(\frac{2x+5}{x^2+5x+6}+2+\frac{-\sin(x+\pi)}{\cos(x+\pi)}\right)$$