Mathematics

Mathematics

Problem

To simplify sin⁡(x+h)−sin⁡xh\dfrac{\sin(x+h)-\sin x}{h} while finding ddx(sin⁡x)\dfrac{d}{dx}(\sin x), which rearrangement correctly prepares the expression for taking the limit h→0h\to 0? a) sin⁡x(1−cos⁡h)+cos⁡xsin⁡hh\displaystyle\frac{\sin x(1-\cos h)+\cos x\sin h}{h} b) sin⁡x(cos⁡h−1)+cos⁡xsin⁡hh\displaystyle\frac{\sin x(\cos h-1)+\cos x\sin h}{h} c) cos⁡x(cos⁡h−1)+sin⁡xsin⁡hh\displaystyle\frac{\cos x(\cos h-1)+\sin x\sin h}{h} d) sin⁡xcos⁡h+cos⁡xsin⁡hh\displaystyle\frac{\sin x\cos h+\cos x\sin h}{h}
Intuitive explanation

This is about recognizing that the difference quotient for sine won't simplify by direct substitution—you need a trig identity to break apart sin⁡(x+h)\sin(x+h) first. The key insight is to use the angle addition formula sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h) = \sin x \cos h + \cos x \sin h, which lets you factor out common terms and separate the limit into pieces you can actually evaluate as h→0h \to 0. Once you expand and rearrange, the path to the derivative becomes clear.

Intuitive explanation

This is about recognizing that the difference quotient for sine won't simplify by direct substitution—you need a trig identity to break apart sin⁡(x+h)\sin(x+h) first. The key insight is to use the angle addition formula sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h) = \sin x \cos h + \cos x \sin h, which lets you factor out common terms and separate the limit into pieces you can actually evaluate as h→0h \to 0. Once you expand and rearrange, the path to the derivative becomes clear.

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