Question

Obtain the following derivatives from first principles: (a) the first derivative of \(3 x+4\) (b) the first, second and third derivatives of \(x^{2}+x ;\) (c) the first derivative of \(\sin x\).

Short answer

a) The first derivative of \(3x + 4\) is 3. b) The first derivative of \(x^2 + x\) is \(2x + 1\), the second derivative is 2, and the third derivative is 0. c) The first derivative of \(\sin x\) is \(\cos x\).

Step-by-step solution

- Understanding first principles

The derivative of a function from first principles is given by the limit definition: \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]

- Derivative of \(3x + 4\)

Let's find the first derivative of the function \(f(x) = 3x + 4\): 1. Substitute \(f(x)\) into the limit definition: \[f'(x) = \lim_{{h \to 0}} \frac{{(3(x+h) + 4) - (3x + 4)}}{h}\] 2. Simplify the expression: \[\lim_{{h \to 0}} \frac{{3x + 3h + 4 - 3x - 4}}{h} = \lim_{{h \to 0}} \frac{{3h}}{h}\] 3. After canceling out \(h\) in the numerator and denominator: \[\lim_{{h \to 0}} 3 = 3\]

- First Derivative of \(x^2 + x\)

Next, let's find the first derivative of \(f(x) = x^2 + x\): 1. Use the limit definition: \[f'(x) = \lim_{{h \to 0}} \frac{{((x+h)^2 + (x+h)) - (x^2 + x)}}{h}\] 2. Simplify the expression: \[\lim_{{h \to 0}} \frac{{(x^2 + 2xh + h^2 + x + h) - (x^2 + x)}}{h} = \lim_{{h \to 0}} \frac{{2xh + h^2 + h}}{h} = \lim_{{h \to 0}} (2x + h + 1)\] 3. Finally: \[ \lim_{{h \to 0}} (2x + h + 1) = 2x + 1 \]

- Second and Third Derivatives of \(x^2 + x\)

Now, find the second derivative by differentiating \(f'(x) = 2x + 1\): 1. The derivative of \(2x + 1\): \[f''(x) = \frac{d}{dx}(2x + 1) = 2\] Next, find the third derivative by differentiating \(f''(x) = 2\): 2. The derivative of \(2\): \[ f'''(x) = \frac{d}{dx}(2) = 0 \]

- Derivative of \(\sin x\)

Lastly, find the derivative of \(f(x) = \sin x\): 1. Use the limit definition: \[f'(x) = \lim_{{h \to 0}} \frac{{\sin(x+h) - \sin(x)}}{h}\] 2. Use the angle addition formula namely \(\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)\): 3. Rewrite and separate terms: \[ \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} = \frac{\sin(x)(\cos(h) - 1)}{h} + \frac{\cos(x)\sin(h)}{h} \] 4. Evaluate the limits: \[\lim_{{h \to 0}} \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \lim_{{h \to 0}} \frac{\sin(h)}{h}\] 5. Using standard limits: \[ = \cos(x) \]

Key concepts

limit definition of a derivative

To understand the derivative, we start with the limit definition. This concept refers to finding the slope of the tangent line to a function at any given point. The formal definition is:
\[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \]
Here, we are taking the limit as the difference in the function values (\(f(x+h) - f(x)\)) divided by the interval \(h\) approaches zero.
This helps us note how the function value changes as we make very small changes in the input. Let’s look at an example to illustrate this.

• Consider the function \(f(x) = 3x + 4\). To find its derivative, apply the limit definition.
• First, substitute \(f(x)\) into the limit definition:
  \[f'(x) = \lim_{{h \to 0}} \frac{{(3(x+h) + 4) - (3x + 4)}}{h} \]
  
• Next, simplify the expression:
  \[ \lim_{{h \to 0}} \frac{{3x + 3h + 4 - 3x - 4}}{h} = \lim_{{h \to 0}} \frac{{3h}}{h} \]
  
• Finally, when \(h\) cancels out in the numerator and denominator:
  \[ \lim_{{h \to 0}} 3 = 3 \]
  It showcases how the limit definition works in practical scenarios.

derivative of a linear function

Understanding the derivative of a linear function, like the earlier example, helps to grasp its simplicity. Linear functions are in the form \(f(x) = ax + b\), where \(a\) and \(b\) are constants.
The derivative of a linear function reflects how fast the output (\(y\)) changes per unit change in the input (\(x\)).

• For \(f(x) = 3x + 4\), the coefficient of \(x\) (which is 3) indicates the rate of change.
  
• The derivative, following steps defined, comes out to be the same coefficient: 3.
  This holds for any linear function; hence:
  \[ \frac{{d}}{{dx}}(ax + b) = a \]
  In essence, the derivative of a linear function is the constant slope of that function.

higher-order derivatives

Higher-order derivatives involve differentiating a function multiple times. These derivatives can provide insights into the curvature and concavity of functions.

• Let's consider \( f(x) = x^2 + x \).
• The first derivative, using the limit process, is \( f'(x) = 2x + 1 \).
  
• For the second derivative, we differentiate the first derivative:
  \[ f''(x) = \frac{d}{dx}(2x + 1) = 2 \]
  
• To get the third derivative, differentiate the second derivative:
  \[ f'''(x) = \frac{d}{dx}(2) = 0 \]
  
  It demonstrates that higher-order derivatives for polynomials of degree 2 or less will eventually result in zero. These higher-order derivatives can help in understanding changes in rates of change.

trigonometric derivatives

Trigonometric functions also have specific derivatives. By using the limit definition, one can arrive at these standard results.

Consider the function \( f(x) = \sin{x} \). To derive it, we use:
\[ f'(x) = \lim_{{h \to 0}} \frac{{\sin{(x+h)} - \sin{x}}}{h} \]
Using the angle addition formula, \( \sin{(x+h)} = \sin{x} \cos{h} + \cos{x} \sin{h} \), let's rewrite:
\[ \frac{\sin{x} \cos{h} + \cos{x} \sin{h} - \sin{x}}{h} = \frac{\sin(x)(\cos(h) - 1)}{h} + \frac{\cos(x) \sin{h}}{h} \]
Evaluating these limits separately:
\[ \lim_{{h \to 0}} \sin{x} \frac{\(\cos{h} - 1\)}{h} + \cos{x} \lim_{{h \to 0}} \frac{\sin(h)}{h} = 0 + \cos{x} = \cos{x} \]

Therefore, the derivative of \( \sin{x} \) is \( \cos{x} \). This approach applies broadly for trigonometric derivatives, vital in understanding changes in periodic functions.

These derivations lay the foundation for other trigonometric functions and their respective changes or transformations.