April 9th, 2009 | 
Author: Knowing how to deal with the derivative of a constant multiplying a function is the key to understanding an important property of differentiable functions called linearity. We can say that the constant passes through the differentiation sign.
This can be understood several ways as the constant is a just a scale factor that multiplies the slope at particular points. This can also be shown directly from the definition of the derivative.
![{\cfrac{d}{dx} [cf(x)]} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{cf(x+\Delta x)-cf(x)}{\Delta x} } \\ = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{c\left(f(x+\Delta x)-f(x)\right)}{\Delta x} } \\ = {c\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x)-f(x)}{\Delta x} } \\ = {c\cfrac{d}{dx} f(x)}](http://l.wordpress.com/latex.php?latex=%7B%5Ccfrac%7Bd%7D%7Bdx%7D%20%5Bcf%28x%29%5D%7D%20%3D%20%7B%5Cmathop%7B%5Clim%20%7D%5Climits_%7B%5CDelta%20x%5Cto%200%7D%20%5Ccfrac%7Bcf%28x%2B%5CDelta%20x%29-cf%28x%29%7D%7B%5CDelta%20x%7D%20%7D%20%5C%5C%20%3D%20%7B%5Cmathop%7B%5Clim%20%7D%5Climits_%7B%5CDelta%20x%5Cto%200%7D%20%5Ccfrac%7Bc%5Cleft%28f%28x%2B%5CDelta%20x%29-f%28x%29%5Cright%29%7D%7B%5CDelta%20x%7D%20%7D%20%5C%5C%20%3D%20%7Bc%5Cmathop%7B%5Clim%20%7D%5Climits_%7B%5CDelta%20x%5Cto%200%7D%20%5Ccfrac%7Bf%28x%2B%5CDelta%20x%29-f%28x%29%7D%7B%5CDelta%20x%7D%20%7D%20%5C%5C%20%3D%20%7Bc%5Ccfrac%7Bd%7D%7Bdx%7D%20f%28x%29%7D&bg=FFFFFF&fg=000000&s=1 "{\cfrac{d}{dx} [cf(x)]} = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{cf(x+\Delta x)-cf(x)}{\Delta x} } \\ = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{c\left(f(x+\Delta x)-f(x)\right)}{\Delta x} } \\ = {c\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(x+\Delta x)-f(x)}{\Delta x} } \\ = {c\cfrac{d}{dx} f(x)}")
Example 1. Find the derivative of
Solution 1. We can use the algebra of logarithms to do some manipulation before we take the derivative.
Now we can apply the new rule and write
From our table of derivatives, the derivative of the natural logarithm is just the reciprocal of x.
Example 2. Find the derivative of
Solution 2. We can use linearity to write the derivative in terms of the power rule.
The linearity of differentiation can now be stated as
![[c_1 f(x) + c_2 g(x)]' = c_1 \cfrac{d}{dx}f(x) + c_2 \cfrac{d}{dx}g(x)](http://l.wordpress.com/latex.php?latex=%20%5Bc_1%20f%28x%29%20%2B%20c_2%20g%28x%29%5D%27%20%3D%20c_1%20%5Ccfrac%7Bd%7D%7Bdx%7Df%28x%29%20%2B%20c_2%20%5Ccfrac%7Bd%7D%7Bdx%7Dg%28x%29%20&bg=FFFFFF&fg=000000&s=1 "[c_1 f(x) + c_2 g(x)]' = c_1 \cfrac{d}{dx}f(x) + c_2 \cfrac{d}{dx}g(x)")
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