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RE: LeoThread 2025-03-10 23:28

in LeoFinance7 months ago

Part 5/6:

\lim_{h \to 0} \frac{\cos h - 1}{h} = 0

]

Thus, substituting these limits back into our expression results in:

[

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

]

Consequently, we arrive at the conclusion that:

[

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

]

Conclusion

Through both graphical interpretation and rigorous proof via the definition of a derivative, we have verified that the derivative of the sine function is indeed the cosine function. This connection not only enhances our understanding of sine and its behavior but also illustrates the beauty of calculus in analyzing functional rates of change.