Problem 7 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Given } f(x)=\sin x \text { , find } f^{\prime}(0), f^{\prime}\left(\frac{\pi}{2}\right) \& f^{\prime}(\pi) \text { by first principles. } $$

Step-by-Step Solution

Answer

The derivatives at the given points are \(f^'(0) = 1\), \(f^'(\pi/2) = 0\) and \(f^'(\pi) = -1\).

Step 1: Write the limit definition of the derivative

The limit definition of the derivative is \(f^'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\). So first, substitute \(f(x) = \sin x\) into this definition.

Step 2: Compute the derivative in terms of h

After substituting we get, \(f^'(x) = \lim_{h\to 0} \frac{\sin(x+h) - \sin x}{h}\). Now it's convenient to use the trigonometric addition formula, which states that \(\sin(a + b) = \sin a \cos b + \cos a \sin b\). Thus, the derivative becomes \(f^'(x) = \lim_{h\to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}\).

Step 3: Simplify further

Then, grouping the terms we get \(f^'(x) = \sin x\lim_{h \to 0}(\frac{\cos h - 1}{h}) + \cos x \lim_{h \to 0}(\frac{\sin h}{h})\). We know that \(\lim_{h\to 0}\frac{\sin h}{h} = 1\) and \(\lim_{h\to 0}\frac{\cos h - 1}{h} = 0\) so we end up with \(f^'(x) = \cos x\). This is the derivative of the function \(f(x) = \sin x\).

Step 4: Substitute each specified value into the derivative

Now, for \(x = 0\), \(f^'(0) = \cos 0 = 1\).\nFor \(x = \pi/2\), \(f^'(\pi/2) = \cos (\pi/2) = 0\). And for \(x = \pi\), \(f^'(\pi) = \cos \pi = -1\).

Key Concepts

Limit Definition of Derivative

The derivative of a function at any point gives us the instantaneous rate of change of the function at that point. It helps us understand how the function behaves as the input changes. To find this derivative using the limit definition, we start with the general formula:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Here, we are essentially finding the slope of the tangent line to the curve at the point \(x\). To do this:

• Substitute the function in question into the formula, which is \(f(x) = \sin x\) in our exercise.
• This substitution leads us to: \(f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}\).

Understanding the limit helps us derive the formula needed to calculate the derivative at any specified point.

Trigonometric Addition Formula

The trigonometric addition formulas are crucial tools in simplifying the expressions we derive from the limit definition. Specifically for our function, we make use of the sine addition formula:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
Using this formula:

• We substitute \(x+h\) to become \(\sin(x) \cos(h) + \cos(x) \sin(h)\).
• Therefore, \(\sin(x+h) - \sin x\) becomes \((\sin x \cos(h) + \cos x \sin(h)) - \sin x\).

These steps allow us to simplify our derivative expression to involve more basic forms like \(\cos h - 1\) and \(\sin h\). Employing these formulas makes it easier to navigate through the derivation.

Sine and Cosine Limits

When simplifying expressions involving trigonometric functions, particularly in derivatives, recognizing key limits can be very useful. Two fundamental limits critical in the context of our function include:

• \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \)
• \( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \)

By knowing these limits:

• The term \(\cos x \lim_{h \to 0} \frac{\sin h}{h} = \cos x \times 1 = \cos x\) is greatly simplified.
• The term \(\sin x \lim_{h \to 0} \frac{\cos h - 1}{h} = \sin x \times 0 = 0\).

Ultimately, these calculate the derivative at any point \(x\) for the sine function, consistently yielding \(f'(x) = \cos x\). Understanding these limits is key to simplifying our derivative calculations.

First Principles in Calculus

First principles refer to the foundational concepts or basic methods in calculus used to find derivatives. This involves using the original definition of the derivative rather than relying on shortcut rules or known derivatives:

• The step-by-step approach involving limits and algebraic manipulation exemplifies the use of first principles.
• It grounds students in understanding how derivatives come about directly from functions themselves, without abstracting away by using preset rules.

Mastering derivatives using first principles enriches one's understanding of calculus, offering insights into why functions behave as they do. By evaluating derivatives of \(f(x) = \sin x\) at points like \(0\), \(\frac{\pi}{2}\), and \(\pi\), students get to explore and comprehend these foundational techniques.