Differentiation
3.0 Properties of Differentiation
3.0 Properties of Differentiation
Sum or Difference: To differentiate a sum or difference, we have to differentiate the individual terms and then put them back together with the appropriate signs. This property is not limited to two functions.$$\frac{d}{{dx}}\left( {f(x) \pm g(x)} \right) = \frac{d}{{dx}}f(x) \pm \frac{d}{{dx}}g(x)$$ Proof: $$\begin{equation} \begin{aligned} {\left( {f\left( x \right) \pm g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right) \pm g\left( {x + h} \right)} \right) - \left( {f\left( x \right) \pm g\left( x \right)} \right)}}{h} \\ {\left( {f\left( x \right) \pm g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right) \pm \left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h} \\ {\left( {f\left( x \right) \pm g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)}}{h} \pm \frac{{\left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h}} \right) \\ {\left( {f\left( x \right) \pm g\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)}}{h} \pm \mathop {\lim }\limits_{h \to 0} \frac{{\left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h} \\ {\left( {f\left( x \right) \pm g\left( x \right)} \right)^\prime } = f'\left( x \right) \pm g'\left( x \right) \\\end{aligned} \end{equation} $$
Example 2. $f(x)=x$ and $g(x)=x^2$ Prove that $\frac{d}{{dx}}\left( {f(x) \pm g(x)} \right) = \frac{d}{{dx}}f(x) \pm \frac{d}{{dx}}g(x)$
Solution: Given: $$f(x)=x \quad and \quad g(x)=x^2$$ $$f(x)+g(x)=x^2+x$$
L.H.S: $$\frac{d}{{dx}}\left( {f(x) \pm g(x)} \right) = \frac{d}{dx}(x^2+x)$$$$\frac{d}{dx}(x^2+x)=\frac{d}{dx}(x^2)+\frac{d}{dx}(x)$$
Since, we know that, $$\frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{dx}(x^2+x)=2x^{2-1}+1x^{1-1}$$$$\frac{d}{dx}(x^2+x)=2x^{1}+x^{0}=2x+1$$
R.H.S: $$\frac{d}{dx}f(x)=\frac{d}{dx}x = 1$$$$\frac{d}{dx}g(x)=\frac{d}{dx}(x^2)=(2)x^{2-1}=2x$$$$ \frac{d}{{dx}}f(x) \pm \frac{d}{{dx}}g(x)=2x+1$$
Thus, L.H.S$=$R.H.S
