In differential calculus, using the first principle (that is, the basic definition of the derivative), what is the derivative of sin x with respect to x?

Introduction / Context:
This question belongs to differential calculus and specifically tests knowledge of the derivative of the sine function. The phrase “using the first principle” refers to the fundamental definition of the derivative as a limit of a difference quotient. In practice, for objective questions it is enough to recall the standard derivative formula, but understanding that it comes from the first principle is conceptually important.

Given Data / Assumptions:
- We are dealing with the function y = sin x.
- The question asks for dy/dx, the derivative of sin x with respect to x.
- “Using the first principle” hints at the limit definition, but the final result must be a standard function of x.
- x is a real variable, measured in radians for trigonometric differentiation.

Concept / Approach:
The derivative of a function f(x) using the first principle is defined as limit as h approaches 0 of [f(x + h) − f(x)] / h. For f(x) = sin x, we substitute sin(x + h) and use trigonometric identities to simplify. Ultimately, standard limit results for sin h / h and (cos h − 1) / h lead to the well known formula that the derivative of sin x is cos x when angles are measured in radians.

Step-by-Step Solution:
Start from the first principle: d/dx(sin x) = limit as h → 0 of [sin(x + h) − sin x] / h. Use the identity sin(x + h) = sin x cos h + cos x sin h. Then numerator = sin x cos h + cos x sin h − sin x. Group terms: sin x(cos h − 1) + cos x sin h. Divide by h: [sin x(cos h − 1)/h] + [cos x sin h / h]. As h → 0, (cos h − 1)/h → 0 and sin h / h → 1. Therefore, the limit becomes 0 + cos x * 1 = cos x.

Verification / Alternative check:
We can check consistency with related derivatives. The derivative of cos x is known to be −sin x. Since sin x and cos x are phase shifted, it is reasonable that the derivative of sin x is cos x. Differentiating sin x numerically at a few points using small increments gives approximate slopes that match cos x values, reinforcing the correctness.

Why Other Options Are Wrong:
Option −sinx is the derivative of cos x, not of sin x, so it is mismatched.
Option −cosx would correspond to the derivative of sin x if there were a negative sign difference in the definition, which is not the case in standard calculus.
Option sinx itself would imply that sin x is its own derivative, similar to exponential functions like e^x, which is not true for trigonometric functions.

Common Pitfalls:
Learners sometimes forget that derivatives of trigonometric functions are valid only if the angle is in radians when using standard formulas. Another pitfall is confusing the derivative pairs, such as mixing up d(sin x)/dx with d(cos x)/dx. Remembering the simple chain sin → cos → −sin → −cos in order helps avoid this confusion.

Final Answer:
The derivative of sin x with respect to x, using the first principle, is cosx.