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Differentiate w.r.t. μ_0
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\frac{\mathrm{d}}{\mathrm{d}\mu _{0}}(\sin(\mu _{0}))=\left(\lim_{h\to 0}\frac{\sin(\mu _{0}+h)-\sin(\mu _{0})}{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+\mu _{0})-\sin(\mu _{0})}{h}
Use the Sum Formula for Sine.
\lim_{h\to 0}\frac{\sin(\mu _{0})\left(\cos(h)-1\right)+\cos(\mu _{0})\sin(h)}{h}
Factor out \sin(\mu _{0}).
\left(\lim_{h\to 0}\sin(\mu _{0})\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\left(\lim_{h\to 0}\cos(\mu _{0})\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Rewrite the limit.
\sin(\mu _{0})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\mu _{0})\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Use the fact that \mu _{0} is a constant when computing limits as h goes to 0.
\sin(\mu _{0})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\mu _{0})
The limit \lim_{\mu _{0}\to 0}\frac{\sin(\mu _{0})}{\mu _{0}} 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_{\mu _{0}\to 0}\frac{\sin(\mu _{0})}{\mu _{0}} 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(\mu _{0})
Substitute the value 0 into the expression \sin(\mu _{0})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\mu _{0}).