解 (1)方程两边求导数得 2yy??2y?2xy??0 ??于是 (y?x)y??y??
y??y? y?x (2)方程两边求导数得
3x2?3y2y??2ay?3axy??0??于是 (y2?ax)y??ay?x2??
y??ay?x2? y2?ax (3)方程两边求导数得 y?xy??ex?y(1?y?)??
于是 (x?ex?y)y??ex?y?y??
ex?y?y? y??x?yx?e (4)方程两边求导数得 y???ey?xeyy??
于是 (1?xey)y???ey?
yy???ey?
1?xe2222?求曲线x3?y3?a3在点(2a, 2a)处的切线方程和法线方程?
44解 方程两边求导数得
2x?3?2y?3y??0? 33于是y????1x3?1311?
y在点(2a, 2a)处y???1?
44所求切线方程为
y?2a??(x?2a)?即x?y?2a?
442所求法线方程为
y?2a?(x?2a)?即x?y?0?
44d2y 3?求由下列方程所确定的隐函数y的二阶导数2?
dx22
(1) x?y?1??
(2) b2x2?a2y2?a2b2? (3) y?tan(x?y)? (4) y?1?xey?
解 (1)方程两边求导数得 2x?2yy??0??y??x?
yy?xxy?xy?yy2?x2x1? y???()??2????yyy2y3y3 (2)方程两边求导数得 2b2x?2a2yy??0??
2by???2?x? ay2bx)y?x(??2y2y?xy?2abb y????2?2??2?ayay22a2y2?b2x24bb??2???23? aa2y3ay (3)方程两边求导数得 y??sec2(x?y)?(1?y?)?
sec2(x?y)1 y???1?sec2(x?y)cos2(x?y)?1sin2(x?y)?cos2(x?y)???1?12? 2?sin(x?y)y2(1?y2)221? y???3y??3(?1?2)??5yyyy (4)方程两边求导数得 y??ey?xeyy???
yyyeeey????? 1?xey1?(y?1)2?yeyy?(2?y)?ey(?y?)ey(3?y)y?e2y(3?y)? y?????223(2?y)(2?y)(2?y) 4?用对数求导法求下列函数的导数?
(1) y?(x)x?
1?x
(2)y?55x?5?
x2?2x?2(3?x)4 (3)y??
(x?1)5 (4)y?xsinx1?ex??解 (1)两边取对数得 ln y?xln|x|?xln|1?x|， 两边求导得
1y??lnx?x?1?ln(1?x)?x?1? yx1?x于是 y??(x)x[lnx?1]?
1?x1?x1?x(2)两边取对数得
lny?1ln|x?5|?1ln(x2?2)?
525两边求导得
1y??1?1?1?2x??y5x?525x2?2于是 y??1555x?5?[1?1?2x]? 22x?55x?2x?2(3)两边取对数得
lny?1ln(x?2)?4ln(3?x)?5ln(x?1)?
2两边求导得
1y??1?4?5? y2(x?2)3?xx?1于是 y??x?2(3?x)4[1?4?5] 52(x?2)x?3x?1(x?1)(4)两边取对数得
lny?1lnx?1lnsinx?1ln(1?ex)?
224两边求导得
1y??1?1cotx?ex? y2x24(1?ex)x1ex1?于是 y?xsinx1?e[?coxt?] 2x24(1?ex)x1ex2?xsinx1?e[?2cotx?x]? 4xe?1 5? 求下列参数方程所确定的函数的导数
?x?at2 (1) ?? 2y?bt?x??(1?sin?) (2) ?? ?y??cos??dyyt?3bt23b解 (1)???t?
dxxt?2at2adyy?(2)???cos???sin??
?1?sin???cos?dxx??x?etsint,?时dy的值? 6?已知?求当t?t3dx?y?ecost.dyyt?etcost?etsintcost?sint 解 ? ???dxxt?etsint?etcostsint?costdy? dx1?3dy当t??时??22?1?3?3?2?
3dx1?31?3227?写出下列曲线在所给参数值相应的点处的切线方程和法线方程?
x?sint? (1) ??y?cos2t?在t?处?
4??x?3at2? (2) ?1?t2? 在t=2处?
?y?3at2?1?tdyy? 解 (1)?t??2sin2t??
dxxt?cost?)?2sin(2?dy4??2??22?x?2?y?0??当t??时??00dx242cos?42所求切线方程为?
y??22(x?2)?即22x?y?2?0?
2所求法线方程为
y??1(x?2)?即2x?4y?1?0?
2?226at(1?t2)?3at2?2t6at? (2)yt???(1?t2)2(1?t2)23a(1?t2)?3at?2t3a?3at2? xt???(1?t2)2(1?t2)2dyyt???6at2?2t2? dxxt?3a?3at1?tdy 当t?2时??2?22??4?x0?6a?y0?12a?
dx1?2355所求切线方程为?
y?12a??4(x?6a)?即4x?3y?12a?0? 535所求法线方程为
y?12a?3(x?6a)?即3x?4y?6a?0? 545d2y8?求下列参数方程所确定的函数的二阶导数2?
dx2?t? (1) ?x?2?
??y?1?t.x?acost (2) ??y?bsint?
??x?3e?t (3) ?? t?y?2e?x?ft(t) (4) ??设f??(t)存在且不为零??ty?tf(t)?f(t)?12dyy?dy(y?)?2解 (1)?t??1?2?xt?t?1?
dxxt?tdxxt?tt3
高等数学上册第六版课后习题答案
