Math  /  Calculus

Question3. 求下列函数的微分: (1) y=1x+2xy=\frac{1}{x}+2 \sqrt{x} (2) y=xsin2xy=x \sin 2 x (3) y=xx2+1y=\frac{x}{\sqrt{x^{2}+1}}; (4) y=ln2(1x)y=\ln ^{2}(1-x) ; (5) y=x2e2xy=x^{2} \mathrm{e}^{2 x}; (6) y=excos(3x)y=\mathrm{e}^{-x} \cos (3-x); (7) y=arcsin1x2y=\arcsin \sqrt{1-x^{2}}; (8) y=tan2(1+2x2)y=\tan ^{2}\left(1+2 x^{2}\right); (9) y=arctan1x21+x2y=\arctan \frac{1-x^{2}}{1+x^{2}} (10) s=Asin(ωt+φ)(A,ω,φs=A \sin (\omega t+\varphi)(A, \omega, \varphi 是常数 )).
4. 将适当的函数填人下列括号内,使等式成立: (1) d()=2 dx\mathrm{d}(\quad)=2 \mathrm{~d} x; (2) d()=3xdxd(\quad)=3 x d x (3) d()=cost dt\mathrm{d}(\quad)=\cos t \mathrm{~d} t (4) d()=sinωxdx(ω0)d(\quad)=\sin \omega x d x \quad(\omega \neq 0); (5) d()=11+x dx\mathrm{d}(\quad)=\frac{1}{1+x} \mathrm{~d} x (6) d()=e2xdxd(\quad)=e^{-2 x} d x (7) d()=1x dxd(\quad)=\frac{1}{\sqrt{x}} \mathrm{~d} x (8) d()=sec23x dxd(\quad)=\sec ^{2} 3 x \mathrm{~d} x.

Studdy Solution
Find the function such that its differential is sec23xdx\sec^2 3x \, dx.
Using integration: sec23xdx=13tan3x+C \int \sec^2 3x \, dx = \frac{1}{3} \tan 3x + C The function is: f(x)=13tan3x+C f(x) = \frac{1}{3} \tan 3x + C

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