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Products and Quotients

We have two very convient rules for finding the derivatives of sums and differences:

$$\frac{d}{dx}(f+g) = \frac{df}{dx} + \frac{dg}{dx}$$


$$\frac{d}{dx}(f-g) = \frac{df}{dx} - \frac{dg}{dx}$$

So, does this work for products and quotients? For example, is it true that

$$\begin{equation}\frac{d}{dx}(fg) = \frac{df}{dx} \cdot \frac{dg}{dx}\end{equation}$$


$$\begin{equation}\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{\frac{df}{dx}}{\frac{dg}{dx}}\end{equation}$$

To come to your own conclusions, try looking at these examples.

Is the derivative of a product, the product of the derivatives?

Consider the functions $f(x) = x^2$ and $g(x) = 1$. Calculate



$$\frac{df}{dx} \cdot \frac{dg}{dx}$$

Then compare your results to see if equation (1) is correct.

Is the derivative of a quotient, the quotient of the derivatives?

Equation (2) is not true. See if you can come up with two functions that demonstrate this. (Don't use the two I used in the previous section. There's a technical reason why you should avoid them that I'll explain below once you're done.)

Hint: Think about two monomials.

One Solution: If $f(x)=x^2$ and $g(x)=x$ then $f'(x)=2x$ and $g'(x)=1$ so $f' / g' = 2x$ but $D_x(f/g) = D_x(x^2/x) = D_x(x) = 1$.

So why shouldn't you use my examples? Those two functions would seem to invalidate the formula since the denominator of $f'/g'$ would be 0. However, it's not uncommon for formulas to include the qualifier, "Where both sides are defined." We saw this, for example, when we were talking about limit properties that involved quotients. For example,

$$\lim_\limits{x\to c}\left(\frac{f}{g}\right) = \frac{\lim_\limits{x\to c}f}{\lim_\limits{x\to c}g}$$

if $\lim_\limits{x\to c}g \ne 0$. So, if you used my two examples, it would be possible that equation (2) was still true just not in the special case where the denominator was 0.

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