13 - Linearity Rule for Differentiation - 2

Let $y=f(x)+g(x)$  , then

$\frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h}$

$=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)+g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}+\underset{h\to 0}{\lim }\frac{g(x+h)-g(x)}{h}$

$=f^{\prime }(x)+g^{\prime }(x)$

Note that if we had $y=f(x)-g(x)$  , then we can consider it as $y=f(x)+(-g(x))$   and we would have

$\frac{dy}{dx}=\frac{d}{dx}(f(x))+\frac{d}{dx}(-g(x))$

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

$=f^{\prime }(x)-g^{\prime }(x)$

We can combine both the cases to say that if $y=f(x)\pm g(x)$  , then $\frac{dy}{dx}=f^{\prime }(x)\pm g^{\prime }(x)$

Basically what this says is that the derivative of the sum or difference of two functions is the sum or difference of the derivatives of the two functions. In fact, we can extend this statement to even three or more functions.

Here are some straight forward examples that employ this differentiation rule.

1. If $y=x^3+x^5$  , then

$\frac{dy}{dx}=\frac{d}{dx}(x^3+x^5)=\frac{d}{dx}(x^3)+\frac{d}{dx}(x^5)=3x^2+5x^4$

2. If $y=x^{-3}+x^3$  , then

$\frac{dy}{dx}=\frac{d}{dx}(x^{-3}+x^3)=\frac{d}{dx}(x^{-3})+\frac{d}{dx}(x^3)=-3x^{-4}+3x^2$

3. If $y=x^2+x+6$  , then $\frac{dy}{dx}=\frac{d}{dx}(x^2+x+6)$

$=\frac{d}{dx}(x^2)+\frac{d}{dx}(x)+\frac{d}{dx}(6)$

$=2x+1+0$

$=2x+1$

4. If $y=3x^2+5x+8$  , then

$\frac{dy}{dx}=\frac{d}{dx}(3x^2)+\frac{d}{dx}(5x)+\frac{d}{dx}(8)$

$=3\frac{d}{dx}(x^2)+5\frac{d}{dx}(x)+\frac{d}{dx}(8)$

$=3(2x)+5(1)+0$

$=6x+5$

5. If $y=5x^3-2x^2-3x+2$  , then

$\frac{dy}{dx}=\frac{d}{dx}(5x^3)-\frac{d}{dx}(2x^2)-\frac{d}{dx}(3x)+\frac{d}{dx}(2)$

$=5\frac{d}{dx}(x^3)-2\frac{d}{dx}(x^2)-3\frac{d}{dx}(x)+\frac{d}{dx}(2)$

$=5(3x^2)-2(2x)-3(1)+0$

$=15x^2-4x-3$