,

RadfordMathematics.com

Online Mathematics Book

Operations with Derivatives

We now learn how to differentiate sums of functions as well as functions multiplied by a constant.
So far we have learnt how to differentiate a function made of one term, but more often than not function are made by adding, or subtracting any linear combination of single terms.
By the end of this section we'll know how to differentiate: \[\alpha f(x)+\beta g(x)\] Where \(f(x)\) and \(g(x)\) are two functions and \(\alpha \) and \(\beta \) are two real numbers.

Differentiating Sums & Differences

Given two functions \(f(x)\) and \(g(x)\), to differentiate their sum, or their difference, we differentiate each of their terms as though they were on their own: \[\begin{pmatrix} f(x) + g(x) \end{pmatrix}' = f'(x)+g'(x) \] and \[\begin{pmatrix} f(x) - g(x) \end{pmatrix}' = f'(x) - g'(x) \]

Example

Differentiate each of the following functions with respect to \(x\):

  1. \(f(x)=4x^5+7x\)

  2. \(y=2x^3-2\sqrt{x}\)

  3. \(f(x)=\frac{3}{x} + x^4-7\)

Solution +

Solution

Using the fact that the derivative of any sum, or difference, of two functions equals to the sum, or difference, of the derivatives of those functions, we differentiate each of the following treating each term as if it were on its own.

  1. The derivative of \(f(x)=4x^5+7x\) is: \[\begin{aligned} f'(x) &= 5\times 4x^{5-1}+7x^{1-1} \\ &= 20x^4+7x^0 \\ f'(x) &= 20x^4+7 \end{aligned}\]

  2. Remembering that \(\begin{pmatrix} \sqrt{x} \end{pmatrix}' = \frac{1}{2\sqrt{x}}\), the derivative of \(y=2x^3-2\sqrt{x}\) is: \[\begin{aligned} \frac{dy}{dx} &= 3\times 2x^{3-1} - 2\times \frac{1}{2\sqrt{x}} \\ &= 6x^2-\frac{2}{2\sqrt{x}} \\ & = 6x^2 - \frac{1}{\sqrt{x}} \end{aligned}\]

  3. Remembering that \(\begin{pmatrix} \frac{1}{x} \end{pmatrix}' = - \frac{1}{x^2}\), we differentiate \(f(x) = \frac{3}{x^2} + x^4 - 7\) as follows: \[ f'(x) = -\frac{3}{x^2} + 4x^{4-1} - 0 = -\frac{3}{x^2} + 4x^{3} \]

Differentiating Functions Multiplied by a Scalar

Given a function \(f(x)\) and a scalar \(\alpha \in \mathbb{R}\) (a real number) the derivative of the product \(\alpha f(x)\) equals to the product of \(\alpha \) and the derivative \(f'(x)\).

That's: \[\begin{pmatrix} \alpha f(x) \end{pmatrix}' = \alpha f'(x)\]

Differentiating Linear Combinations of Functions

Given two functions \(f(x)\) and \(g(x)\) as well as two scalars \(\alpha\) and \(\beta \) both in \(\mathbb{R}\), the derivative of any linear combination of \(f(x)\) and \(g(x)\) equals to the same linear combination of their derivatives.
That's: \[\begin{pmatrix} \alpha f(x) + \beta g(x) \end{pmatrix}' = \alpha f'(x) + \beta g'(x)\]

Example

Given \(f(x)=x^3\), \(g(x)=\sqrt{x}\) and \(h(x) = \frac{1}{x^2}\), differentiate each of the following:

  1. \(3f(x)+4g(x)\)

  2. \(g(x)-6h(x)\)

  3. \(\frac{5}{6}f(x)+\frac{1}{8}h(x)-\frac{3}{4}g(x)\)

Solution +

Solution

  1. Remembering that \(\frac{1}{x^2} = x^{-2}\) so \(\begin{pmatrix} \frac{1}{x^2} \end{pmatrix}'= -2x^{-2-1} = -2x^{-3}\), we differentiate as follows: \[\begin{aligned} \begin{pmatrix} 3f(x)+4g(x) \end{pmatrix}' &= 3f'(x) +4g'(x) \\ & = 3\times 3x^{3-1} + 4\times (-2)x^{-2-1} \\ & = 9x^2 - 8 x^{-3} \\ \begin{pmatrix} 3f(x)+4g(x) \end{pmatrix}' & = 9x^2-\frac{8}{x^3} \end{aligned}\]

  2. Remembering that \(\begin{pmatrix} \sqrt{x} \end{pmatrix}' = \frac{1}{2\sqrt{x}}\), we differentiate as follows: \[\begin{aligned} \begin{pmatrix} g(x)-6h(x) \end{pmatrix}' &= g'(x) - 6h'(x) \\ & = \frac{1}{2\sqrt{x}} - 6\times (-2)x^{-2-1} \\ & = \frac{1}{2\sqrt{x}} + 12x^{-3} \\ \begin{pmatrix} g(x)-6h(x) \end{pmatrix}' &= \frac{1}{2\sqrt{x}} + \frac{12}{x^3} \end{aligned}\]

  3. To differentiate \(\frac{5}{6}f(x)+\frac{1}{8}h(x)-\frac{3}{4}g(x)\), we use all that we have seen here to write: \[\begin{aligned} \begin{pmatrix} \frac{5}{6}f(x)+\frac{1}{8}h(x)-\frac{3}{4}g(x) \end{pmatrix}' &= \frac{5}{6}f'(x)+\frac{1}{8}h'(x)-\frac{3}{4}g'(x) \\ &= 3\times \frac{5}{6} x^{3-1} -2\times \frac{1}{8}x^{-2-1} - \frac{3}{4} \times \frac{1}{2}\sqrt{x} \\ & = \frac{15}{6}x^2-\frac{2}{8}x^{-3} - \frac{3}{8\sqrt{x}} \\ & = \frac{15}{6}x^2-\frac{1}{4}\frac{1}{x^3} - \frac{3}{8\sqrt{x}} \\ \begin{pmatrix} \frac{5}{6}f(x)+\frac{1}{8}h(x)-\frac{3}{4}g(x) \end{pmatrix}' & = \frac{5}{2}x^2-\frac{1}{4x^3} - \frac{3}{8\sqrt{x}} \end{aligned}\]