$$\cos x =$$
$-\sin(x)$
$$\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))=\left(\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{h}\right)$$
$$\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{h}$$
$$\lim_{h\to 0}\frac{\cos(x)\left(\cos(h)-1\right)-\sin(x)\sin(h)}{h}$$
$$\left(\lim_{h\to 0}\cos(x)\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\left(\lim_{h\to 0}\sin(x)\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$
$$\cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$
$$\cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x)$$
$$\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)=\left(\lim_{h\to 0}\frac{\left(\cos(h)-1\right)\left(\cos(h)+1\right)}{h\left(\cos(h)+1\right)}\right)$$
$$\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}$$
$$\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}$$
$$\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$
$$-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$
$$\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)=0$$
$$-\sin(x)$$
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$\cos(x)$
