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