Find out the derivtive of (cosx)/x using the first principle of derivative.

To find the derivative of the function \( f(x) = \frac{\cos x}{x} \) using the first principle of derivatives, we will apply the definition of the derivative. This principle states that the derivative of a function at a point is the limit of the average rate of change of the function as the interval approaches zero. In mathematical terms, the derivative \( f'(x) \) is defined as:

Definition of the Derivative

The derivative of \( f(x) \) at a point \( x \) is given by:

\( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Applying the Definition

For our function \( f(x) = \frac{\cos x}{x} \), we first need to compute \( f(x+h) \):

\( f(x+h) = \frac{\cos(x+h)}{x+h} \)

Now, substituting \( f(x) \) and \( f(x+h) \) into the derivative formula, we have:

\( f'(x) = \lim_{h \to 0} \frac{\frac{\cos(x+h)}{x+h} - \frac{\cos x}{x}}{h} \)

Finding a Common Denominator

To simplify the expression, we need a common denominator for the two fractions in the numerator:

\( f'(x) = \lim_{h \to 0} \frac{x \cos(x+h) - (x+h) \cos x}{h(x+h)} \)

Expanding the Numerator

Next, we will expand the numerator:

• Distributing \( (x+h) \) gives us \( x \cos(x+h) - x \cos x - h \cos x \).

Thus, we can rewrite the limit as:

\( f'(x) = \lim_{h \to 0} \frac{x (\cos(x+h) - \cos x) - h \cos x}{h(x+h)} \)

Separating the Limit

We can separate this limit into two parts:

\( f'(x) = \lim_{h \to 0} \left( \frac{x (\cos(x+h) - \cos x)}{h(x+h)} - \frac{\cos x}{x+h} \right) \)

Using the Limit of Cosine Difference

As \( h \) approaches 0, the term \( \frac{\cos(x+h) - \cos x}{h} \) approaches \( -\sin x \) (from the derivative of cosine). Therefore, we can substitute this into our limit:

\( f'(x) = \frac{x (-\sin x)}{x^2} - \frac{\cos x}{x} \)

Simplifying the Expression

Now, simplifying gives us:

\( f'(x) = -\frac{\sin x}{x} - \frac{\cos x}{x^2} \)

Final Result

Thus, the derivative of \( f(x) = \frac{\cos x}{x} \) is:

\( f'(x) = -\frac{\sin x}{x} - \frac{\cos x}{x^2} \)

This result shows how the first principle of derivatives can be applied to find the derivative of a quotient of functions, illustrating the power of limits in calculus. If you have any further questions about this process or any other topic, feel free to ask!