3.3 Differentiation rules  (Page 3/14)

We now apply the sum law for limits and the definition of the derivative to obtain

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Applying the constant multiple rule

Find the derivative of $g\left(x\right)=3x^2$ and compare it to the derivative of $f\left(x\right)=x^2.$

We use the power rule directly:

$g^{\prime }\left(x\right)=\frac{d}{dx}\left(3x^2\right)=3\frac{d}{dx}\left(x^2\right)=3\left(2x\right)=6x.$

Since $f\left(x\right)=x^2$ has derivative $f^{\prime }\left(x\right)=2x,$ we see that the derivative of $g\left(x\right)$ is 3 times the derivative of $f\left(x\right).$ This relationship is illustrated in [link] .

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The product rule

Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule    does not follow this pattern. To see why we cannot use this pattern, consider the function $f\left(x\right)=x^2,$ whose derivative is $f^{\prime }\left(x\right)=2x$ and not $\frac{d}{dx}\left(x\right)·\frac{d}{dx}\left(x\right)=1·1=1.$

Proof

We begin by assuming that $f(x)$ and $g(x)$ are differentiable functions. At a key point in this proof we need to use the fact that, since $g(x)$ is differentiable, it is also continuous. In particular, we use the fact that since $g(x)$ is continuous, $\underset{h\to 0}{\text{lim}}g\left(x+h\right)=g(x).$

By applying the limit definition of the derivative to $j\left(x\right)=f\left(x\right)g\left(x\right),$ we obtain