How we can prove d/dx(sin x) is equal to cos x PLZ, GIVE THE TOTAL PROOF FOR IT ….….….….….….….….….….….….….….….…..????????????????????????????????????????????????????????

To prove that the derivative of the sine function is equal to the cosine function, we can use the limit definition of the derivative. This approach not only establishes the relationship but also enhances our understanding of how derivatives work in general.

Understanding the Derivative

The derivative of a function at a certain point gives us the slope of the tangent line to the function at that point. Mathematically, the derivative of a function \( f(x) \) is defined as:

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

Applying the Limit Definition

For our specific case, we want to find the derivative of \( \sin x \). We substitute \( f(x) \) with \( \sin x \):

\( \frac{d}{dx}(\sin x) = \lim_{{h \to 0}} \frac{\sin(x + h) - \sin x}{h} \)

Using the Sine Addition Formula

Next, we can utilize the sine addition formula, which states that:

\( \sin(a + b) = \sin a \cos b + \cos a \sin b \)

By applying this to our expression, we have:

\( \sin(x + h) = \sin x \cos h + \cos x \sin h \)

Substituting Back into the Limit

Now, substituting this expression back into our derivative formula gives us:

\( \frac{d}{dx}(\sin x) = \lim_{{h \to 0}} \frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h} \)

Rearranging this, we can factor out \( \sin x \):

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

Breaking Down the Limit

This expression can be split into two separate limits:

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

Next, we need to evaluate the two limits separately:

• Limit of \( \frac{\sin h}{h} \): It is well-known that \( \lim_{{h \to 0}} \frac{\sin h}{h} = 1 \).
• Limit of \( \frac{\cos h - 1}{h} \): This limit evaluates to 0 as \( h \) approaches 0.

Final Evaluation

Substituting back these evaluations into our expression gives us:

\( = \sin x \cdot 0 + \cos x \cdot 1 \)

Thus, we find that:

\( \frac{d}{dx}(\sin x) = \cos x \)

Summary of the Proof

In summary, using the limit definition of the derivative and the sine addition formula, we have shown that:

\( \frac{d}{dx}(\sin x) = \cos x \)

This proof not only confirms the derivative relationship but also illustrates fundamental concepts in calculus that will be vital as you continue your studies. If you have any more questions or need clarification on any of the steps, feel free to ask!