微分積分学１ No.7 2005. 6. 8
2.5 双曲線関数（解答） 担当：市原
問題15 次の関数を微分しなさい.
(1)y=−2 tanh(3x) y0=−2×(3x)0× 1
cosh2(3x) =− 6 cosh2(3x)
(2)y= arcsinx·coshx y0= (arcsinx)0×coshx+ arcsinx×(coshx)0
= 1
√1−x2 ×coshx+ arcsinx×sinhx
= coshx
√1−x2 + arcsinx·sinhx
(3)y= sinh(x+ 5)4 y0 =¡
(x+ 5)4¢0
×cosh(x+ 5)4= 4(x+ 5)3cosh(x+ 5)4
(4)y= log(cosh(3x−2)) y0= (cosh(3x−2))0× 1
cosh(3x−2)
= (3x−2)0×sinh(3x−2)× 1 cosh(3x−2)
= 3× sinh(3x−2)
cosh(3x−2) = 3 tanh(3x−2)
(5)y= arctanh (x2+ 3) y0 = arctanh (x2+ 3) =¡
(x2+ 3)¢0
× 1
1−(x2+ 3)2
= 2x× 1
1−(x2+ 3)2 = 2x 1−(x2+ 3)2
= 2x
(1 + (x2+ 3))(1−(x2+ 3)) =− 2x (x2+ 4)(x2+ 2)
(6)y=x(arcsinhx) y0=x(arcsinhx) =x0×arcsinhx+x×(arcsinhx)0
= arcsinhx+x× 1
√x2+ 1 = arcsinhx+ x
√x2+ 1
(7)y= (arcsinh (3x+ 1))2 y0= (arcsinh (3x+ 1))0×2(arcsinh (3x+ 1))1
= (3x+ 1)0× 1
p1 + (3x+ 1)2 ×2 arcsinh (3x+ 1)
= 3× 1
√9x2+ 6x+ 2 ×2 arcsinh (3x+ 1)
= 6 arcsinh (3x+ 1)
√9x2+ 6x+ 2
