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Differentiate w.r.t. ε
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\frac{\mathrm{d}}{\mathrm{d}\epsilon }(\sin(\epsilon ))=\left(\lim_{h\to 0}\frac{\sin(\epsilon +h)-\sin(\epsilon )}{h}\right)
For a function f\left(x\right), the derivative is the limit of \frac{f\left(x+h\right)-f\left(x\right)}{h} as h goes to 0, if that limit exists.
\lim_{h\to 0}\frac{\sin(h+\epsilon )-\sin(\epsilon )}{h}
Use the Sum Formula for Sine.
\lim_{h\to 0}\frac{\sin(\epsilon )\left(\cos(h)-1\right)+\cos(\epsilon )\sin(h)}{h}
Factor out \sin(\epsilon ).
\left(\lim_{h\to 0}\sin(\epsilon )\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\left(\lim_{h\to 0}\cos(\epsilon )\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Rewrite the limit.
\sin(\epsilon )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\epsilon )\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Use the fact that \epsilon is a constant when computing limits as h goes to 0.
\sin(\epsilon )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\epsilon )
The limit \lim_{\epsilon \to 0}\frac{\sin(\epsilon )}{\epsilon } is 1.
\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)=\left(\lim_{h\to 0}\frac{\left(\cos(h)-1\right)\left(\cos(h)+1\right)}{h\left(\cos(h)+1\right)}\right)
To evaluate the limit \lim_{h\to 0}\frac{\cos(h)-1}{h}, first multiply the numerator and denominator by \cos(h)+1.
\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}
Multiply \cos(h)+1 times \cos(h)-1.
\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}
Use the Pythagorean Identity.
\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)
Rewrite the limit.
-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)
The limit \lim_{\epsilon \to 0}\frac{\sin(\epsilon )}{\epsilon } is 1.
\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)=0
Use the fact that \frac{\sin(h)}{\cos(h)+1} is continuous at 0.
\cos(\epsilon )
Substitute the value 0 into the expression \sin(\epsilon )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\epsilon ).