Cách làm cho bạn:
a) \(y = \tan^2 x\)
\(dy = d(\tan^2 x) = (\tan^2 x)'dx = 2\tan x.(\tan x)'dx = \frac{2\tan x}{\cos^{2}x}dx\).
b) \(y = \frac{\cos x}{1-x^{2}}\)
\(dy = d \left ( \frac{\cos x}{1-x^{2}} \right )= \left ( \frac{\cos x}{1-x^{2}} \right )'dx \)
\(= \frac{(\cos x)'.(1-x^{2})-\cos x(1-x^{2})'}{(1-x^{2})^{2}}dx\)
\(= \frac{(x^{2}-1).\sin x+2x\cos x}{(1-x^{2})^{2}}dx\).
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