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I'm home teaching myself calculus because I'm 16 and therefore too young to take an actual class with a teacher, so I apologise if this seems simple.

I understand the definition of the Product Rule and its formula:

"If a function $h(x)=f(x)\times g(x)$ is the product of two differentiable functions $f(x)$ and $g(x)$, then $h'(x) = f(x)\times g'(x)+f'(x)\times g(x)$".

I did a question to find the derivative of $g(x) = (2x+1)(x+4)$ using the Product Rule.

Now on the solutions sheet it says I must begin by writing:

$g'(x)=(2x+1){\bf (1)}+{\bf (2)}(x+4)$

What confuses me are the terms that I have put in bold. (the terms $(1)$ and $(2)$). I believe the term $(1)$ is $g'(x)$ from the formula and the term $(2)$ is $f'(x)$ from the formula.

How am I supposed to know these 2 terms? Am I supposed to find the derivative of $(2x+1)$ and $(x+4)$ before going on to the question?

I also apologise if this is quite messy.

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Note that using the index rule you can easily determine that the derivative of $(x+4)$ is $1$, and the derivative of $(2x+1)$ is $2$. This is where the terms come from. – Shaktal Jul 13 '12 at 21:32
up vote 2 down vote accepted

I don't really understand what the stars are suppose to mean, but I can try to explain the product rules.

As you pointed out,

$h'(x) = f'(x)g(x) + f(x)g'(x)$.

In your example, the function $h(x) = (2x +1)(x + 4)$. First, you need to recognize that $h$ is the product of two functions:

$f(x) = 2x + 1$

$g(x) = x + 4$

In order to apply the product rule formula, you need to find $f'(x)$ and $g'(x)$. I assume that you know how to do this for the two functions above. Hence you should obtain

$f'(x) = 2$

$g'(x) = 1$

Sticking all of these terms into the product the rule

$h'(x) = f'(x)g(x) + f(x)g'(x)$

$h'(x) = (2)(x + 4) + (2x + 1)(1) = 2x + 8 + 2x + 1 = 4x + 9$

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I was trying to make them bold so as to say "these are the parts I'm confused about" – ODP Jul 13 '12 at 21:39
@OllyPrice The product rule assumes that you know how to find $f'(x)$ and $g'(x)$. For example, the $f(x)$ and $g(x)$ are polynomial functions, so you can compute their derivative $2$ and $1$ respectively using the power rules for polynomials or just plug these function into the definition of the derivative $\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$. – William Jul 13 '12 at 21:43

You’re supposed to find the derivatives of $2x+1$ and $x+4$ as part of finding the derivative of their product:

$$\begin{align*} \Big((2x+1)(x+4)\Big)'&=(2x+1)(x+4)'+(x+4)(2x+1)'\\ &=(2x+1)(1)+(x+4)(2)\\ &=(2x+1)+(2x+8)\\ &=4x+9\;. \end{align*}$$

Note that you can check this by multiplying out the original function to get $2x^2+9x+4$ and taking its derivative directly.

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So you have $(2x+1)(x+4) = h(x) = f(x)g(x)$ where $f(x) = 2x + 1$ and $g(x) = x + 4$. So you find the derivative of both $f(x)$ and $g(x)$: $$\begin{align} f'(x) &= 2 \\ g'(x) &= 1. \end{align} $$ And so $$ h'(x) = f'(x)g(x) + f(x)g'(x) = 2(x+4) + (2x+1)1. $$ It seems like the $**$ you have above are supposed to be unknowns - like a fill in the blank.

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Just to add to and perhaps simplify the answers given above, the apostrophe in $f'(x)$ in the formula for the product rule means "take the derivative of $f(x)$." When reading $f'(x)$, we say "f prime of x." So you did in fact answer your own question and the others here have added more detail about how it works. Good luck with your studies!

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Yeah I managed to get that and thanks, I've managed to get through limits and continuity and I'm now onto derivatives so fingers crossed because as far as I know it gets a lot harder??? – ODP Jul 13 '12 at 22:37
@OllyPrice From my experience tutoring college students in calculus courses, the concepts of calculus are relatively easy for students to understand. In homework problems, the algebra used to simplify the answer trips up more people than the actual calculus. Good luck! – Code-Guru Jul 13 '12 at 22:44
Yeah I've found that lots, in America what's the standard age to be taught full-blown calculus? – ODP Jul 13 '12 at 23:05
@OllyPrice I took a calculus class in high school during my senior year when I was 17. I believe that is still about "standard" in the American public education system. There may be some schools that teach it sooner. – Code-Guru Jul 13 '12 at 23:14
@OllyPrice I don't know what "full-blown calculus" means, but in the United States there are AP Calculus classes which are taught generally to student in grade 10, 11, or 12. However rigorous calculus involving proofs are usually only seen in math classes oriented toward math major in college. – William Jul 13 '12 at 23:19

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