3.1 Scalar Differentiation:

  Differentiation

$$\frac{d}{dx} (c(f(x)) = c \frac{d}{dx}f(x)$$ Proof: $$\begin{equation} \begin{aligned} {\left( {cf\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {cf\left( {x + h} \right)} \right) - \left( {cf\left( x \right)} \right)}}{h} \\ {\left( {cf\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{c\left( {f\left( {x + h} \right) - f\left( x \right)} \right)}}{h} \\ {\left( {cf\left( x \right)} \right)^\prime } = c\mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)}}{h} \\ {\left( {cf\left( x \right)} \right)^\prime } = cf'\left( x \right) \\\end{aligned} \end{equation} $$

Example 3. $f(x)=5x^2+4x-3$ and $c= 2$. Prove $\frac{d}{dx} (c(f(x)) = c \frac{d}{dx}f(x)$

Solution: Given: $$f(x)=5x^2+4x-3 \quad c= 2$$ $$c f(x) =2(5x^2+4x-3)=10x^2+8x-6$$

L.H.S: $$\frac{{d}}{{dx}}\left( c f(x) \right) = \frac{d}{dx} (10x^2+8x-6)$$.
Using sum and difference property we get,
$$\frac{{d}}{{dx}}(10x^2+8x-6) =\frac{d}{dx}10x^2 + \frac{d}{dx}8x - \frac{d}{dx}6 $$
Since, we know that $$\frac{d}{dx}x^n = nx^{n-1}$$$$\frac{d}{dx}k=0 \quad where \; k = \; constant $$
$$\frac{{d}}{{dx}}(10x^2+8x-6) =10(2)x^{2-1} + 8x^{1-1} -0 $$$$\frac{{d}}{{dx}}(10x^2+8x-6) =20x+8$$
R.H.S: $$c\frac{d}{{dx}}f(x) = 2\frac{d}{{dx}}\left( {5{x^2} + 4x - 3} \right)$$

Using Sum and Difference and scalar property

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

Since we know that,

$$\frac{d}{{dx}}{x^n} = n{x^{n - 1}}$$$$\frac{d}{{dx}}k = 0\quad where\,k = constant$$$$2\frac{d}{{dx}}\left( {5{x^2} + 4x - 3} \right) = 2\left( {5(2){x^{2 - 1}} + 4{x^{1 - 1}} - 0} \right)$$$$2\frac{d}{{dx}}\left( {5{x^2} + 4x - 3} \right) = 2\left( {10x + 4} \right) = 20x + 8$$

thus, L.H.S = R.H.S