Question

найдите производные с объяснением решения

Answer 1

$59. \ \ y'=((\frac{t^2}{t^3-4t})^3)'=((\frac{t^2}{t\cdot (t^2-4)})^3)'=((\frac{t}{t^2-4})^3)'=3\cdot (\frac{t}{t^2-4})^2 \ \cdot \ \frac{t'\cdot(t^2-4)-t\cdot(t^2-4)'}{(t^2-4)^2}= \\ \\ =\frac{3t^2}{(t^2-4)^2} \cdot\frac{1\cdot (t^2-4)-t\cdot 2t}{(t^2-4)^2}=\frac{3t^2\cdot (t^2-4-2t^2)}{(t^2-4)^4}=\frac{3t^2\cdot (-t^2-4)}{(t^2-4)^4}=-\frac{3t^2\cdot(t^2+4)}{(t^2-4)^4}$

$39. \ \ h'(x)=(x\cdot \tan{(2\sqrt{x}})+7)'=(x\cdot \tan{(2\sqrt{x}}))'+(7)'= \\ \\ = x'\cdot \tan{(2\sqrt{x})} +x\cdot (\tan{(2\sqrt{x})})'+0=\tan{(2\sqrt{x})} +x\cdot \frac{(2\sqrt{x})'}{\cos^2{(2\sqrt{x})}} = \\ \\ = \tan{(2\sqrt{x})} +x\cdot \frac{2\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}}{\cos^2{(2\sqrt{x})}} = \tan{(2\sqrt{x})} +\frac{x\cdot x^{-\frac{1}{2}}}{\cos^2{(2\sqrt{x})}} = \tan{(2\sqrt{x})} +\frac{x^{\frac{1}{2}}}{\cos^2{(2\sqrt{x})}} =$

$= \tan{(2\sqrt{x})} +\frac{\sqrt{x}}{\cos^2{(2\sqrt{x})}}$