问三道微积分题目 急find first derivative x^3-xy+y^2=4f(x)=sec^2tanxf(
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问三道微积分题目 急
find first derivative
x^3-xy+y^2=4
f(x)=sec^2tanx
f(x)=ln(x e^7x)
find first derivative
x^3-xy+y^2=4
f(x)=sec^2tanx
f(x)=ln(x e^7x)
郭敦顒回答:
(x^3-xy+y^2)′=3x^2-(xy)′+2y
∵(xy)′= y +x
∴(x^3-xy+y^2)′=3x^2-x+y.
f(x) =sec^2tanx,应表为f(x)=sec^2xtanx,于是
f(x)′=(sec^2xtanx)′=(sec^2x)′tanx+sec^2x(tanx)′
=2secx(secx)′tanx+sec^2xsec^2x
=2sec^2xtan^2x+1/cos^4x.
f(x)=ln(x e^7x)
f(x)′=[ln(x e^7x)]′=1/(x e^7x)′=1/[ e^7x+ x e^7x(7x)′]
=1/(e^7x+ 7x e^7x)
(x^3-xy+y^2)′=3x^2-(xy)′+2y
∵(xy)′= y +x
∴(x^3-xy+y^2)′=3x^2-x+y.
f(x) =sec^2tanx,应表为f(x)=sec^2xtanx,于是
f(x)′=(sec^2xtanx)′=(sec^2x)′tanx+sec^2x(tanx)′
=2secx(secx)′tanx+sec^2xsec^2x
=2sec^2xtan^2x+1/cos^4x.
f(x)=ln(x e^7x)
f(x)′=[ln(x e^7x)]′=1/(x e^7x)′=1/[ e^7x+ x e^7x(7x)′]
=1/(e^7x+ 7x e^7x)
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