Правила нахождения производных

\(\frac{\triangle f}{\triangle x}при\ \triangle x\rightarrow0\)

\(Число,\ к\ которому\ стремится\frac{\triangle f}{\triangle x}при\ \triangle x\rightarrow0\)

\(\frac{\triangle f}{x}при\ x\rightarrow0\)

\(Число,\ к\ которому\ стремится\frac{\triangle f}{\triangle x}при\ \triangle f\rightarrow0\)

\(\left(st\right)'=s't+t's\)

\(\left(st\right)'=st'+ts'\)

\(\left(st\right)'=t'\cdot s'\)

\(\left(st\right)'=s't\cdot t's\)

\(\left(\frac{h}{p}\right)'=\frac{p'h-h'p}{p^2}\)

\(\left(\frac{h}{p}\right)'=\frac{h'p-p'h}{p^2}\)

\(\left(\frac{h}{p}\right)'=\frac{h'p-p'h}{h^2}\)

\(\left(\frac{h}{p}\right)'=\frac{h'}{p'}\)

\(\frac{1}{2\sqrt{x}}\)

\(\frac{1}{4\sqrt{x}}\)

\(\frac{1}{\sqrt{x}}\)

\(-\frac{1}{\sqrt{x}}\)

\(-\frac{1}{5x^2}\)

\(-\frac{5}{x^2}\)

\(\frac{5}{x^2}\)

5

\(2\cos2x\)

\(2\cos x\)

\(\cos2x\)

\(-2\cos2x\)

\(-\frac{\sin\frac{x}{4}}{4}\)

\(-\frac{\sin x}{4}\)

\(\frac{\sin\frac{x}{4}}{4}\)

\(-4\sin\frac{x}{4}+\pi\)

\(\operatorname{ctg}x-\frac{2}{x^4}-\frac{\sqrt{x}}{2}+2x^3+\frac{\pi}{6}\)

\(\operatorname{ctg}x+\frac{2}{x^4}-\sqrt{x}+2x^3+\frac{\pi}{2}\)

\(\frac{2}{x^4}-\sqrt{x}-\operatorname{ctg}x+3x^2+\frac{\pi}{2}\)

\(tgx-\frac{2}{x^4}-\frac{\sqrt{x}}{2}+2x^3+\frac{\pi}{2}\)

\(\left(\frac{\cos3x}{^{x^4}}\right)'=-\frac{3x\sin x+4\cos x}{x^5}\)

\(\left(\frac{\cos3x}{^{x^4}}\right)'=-\frac{3x\sin3x+4\cos3x}{x^8}\)

\(\left(\frac{\cos3x}{^{x^4}}\right)'=-\frac{3\sin3x}{4x^5}\)

\(\left(\frac{\cos3x}{^{x^4}}\right)'=-\frac{3x\sin3x+4\cos3x}{x^5}\)

\(\left(\sqrt{x}\sin x\right)'=\frac{\sin x+2x\cos x}{2\sqrt{x}}\)

\(\left(\sqrt{x}\sin x\right)'=\frac{1}{2\sqrt{x}}+\cos x\)

\(\left(\sqrt{x}\sin x\right)'=\frac{\sin x+2\sqrt{x}\cos x}{2\sqrt{x}}\)

\(\left(\sqrt{x}\sin x\right)'=\frac{1}{2\sqrt{x}}\cdot\cos x\)