Review Handout for Derivatives and u-Substitution - Math 121

Review Handout for Derivatives and u-Substitution-Math 121 Compute the following derivatives. Do not worry about simplifications here. 1. y = sin3 (x3 ) y ′ = 3 sin2 (x3 ) · cos(x3 ) · 3x2 = 9x2 sin2 (x3 ) cos(x3 ) 2. y = cos2 (3x) y ′ = 2 cos(3x) · (− sin(3x)) · 3 = −6 cos(3x) sin(3x) 3. f (t) = t2 sin5 (2t) f ′ (t) = t2 [5 sin4 (2t) · cos(2t) · 2] + sin5 (2t) · 2t = 10t2 sin4 (2t) cos(2t) + 2t sin5 (2t)    2 4 20  2 4 2 5 H ′ (x) = 5 1 − 2 · (4x−3 ) = 3 1 − 2 4. H(x) = 1 − 2 x x x x p 3 x3 + 8 5. f (x) = 1 x2 f ′ (x) = (x3 + 8)−2/3 · 3x2 = 3 3 (x + 8)2/3   1 + tan t 6. g(t) = 1 + t2 h (1 + t2 ) 3t2 + sec2 g ′ (t) = t3 ... = · −
i 1 t2 t2 ) 2  − t3 + tan 1 t
 · (2t) (1 + t2 )2 1 (−2x + 3)5 8. r(x) = (2x + 1)3 (3x + 1)4 =
(1 +     t4 + 3t2 − (1 + t−2 ) sec2 1t − 2t tan 1t 7. p(x) = r′ (x) = 1 t p′ (x) = −5(−2x + 3)−6 · (−2) = 10 (−2x + 3)−6 (3x + 1)4 · 3(2x + 1)2 · (2) − (2x + 1)3 · 4(3x + 1)3 · (3) (3x + 1)8 6(2x + 1)2 (3x + 1)3 [(3x + 1) − 2(2x + 1)] 6(2x + 1)2 [3x + 1 − 4x − 2] = (3x + 1)8 (3x + 1)5 6(2x + 1)2 (−x − 1) −6(x + 1)(2x + 1)2 = (3x + 1)5 (3x + 1)5   1 + 2x 4 9. S(x) = 1 + 3x  1 + 2x 3 h (1 + 3x) · 2 − (1 + 2x) · 3 i · S ′ (x) = 4 1 + 3x (1 + 3x)2 = = 4(1 + 2x)3 · [2 + 6x − 3 − 6x] −4(1 + 2x)3 = (1 + 3x)5 (1 + 3x)5 10. g(x) = cos(3x) sin(4x)
g ′ (x) = cos(3x) cos(4x) · 4 + sin(4x) − sin(3x) · (3) = 4 cos(3x) cos(4x) − 3 sin(3x) sin(4x) 1 11. g(x) = cos(3x) sin(4x)
sin(4x) − sin(3x) · 3 − cos(3x) cos(4x) · 4 −3 sin(3x) sin(4x) − 4 cos(3x) cos(4x) = g (x) = 2 sin (4x) sin2 (4x) ′
− 5 12. y = (x2 + 3x)4 + x 7 −12/7 5 [4(x2 + 3x)3 (2x + 3) + 1] y ′ = − (x2 + 3x)4 + x 7 ************************************************************************* Compute the following integrals. 1. Z π 9 tan(3x) dx = π 18 1 =− 3 ln 1 √2 3 2 ln 3 = Here 6 or 2. Z e4 e 3. e ex sin(3x) 1 dx = − cos(3x) 3 π 18 1 =− 3  ln  Z dx = Z 4 1 1 √ du = 3 u and  x2 ln x e e +C Z 4 u = 1 + e2x du = 2e2x dx 1 2x 2 du = e dx and π x = 18 x = π9 √ 1 u− 2 du = 6 u 1 e dx = e Here 1 2 ln 3 Here √ 3 2 1 1 du = − ln |u| u 3 1 2 √ 3 2 1 =− 3   1 − ln ln 2 √ !! 3 2 √ √  √  ln 3 1 1 = − ln 1 − ln 3 = − 0 − ln 3 = 3 3 3 =⇒ u = cos =⇒ u = cos 4 =6 1 Z xe and e dx = 2 x2 u = x2 du = 2x dx du = x dx Z e2x 1 10 1 1 4. dx = du = ln |u| 2x 1+e 2 2 u 2 0    10 ln 5 1 ln = = 2 2 2 Z 1 2 √ 4− 3π 18
3π 9
= cos = cos π 6
π 3
= = √ 3 2 1 2 √  1 = 6(2 − 1) = 6 x = e =⇒ u = ln e = 1 x = e4 =⇒ u = ln e4 = 4 2 +1 2 1 √ 3 Z u = cos(3x) du = − sin(3x)(3) dx − 13 du = sin(3x) dx u = ln x du = x1 dx x2 +ln x+1 or = π 9 !! 3 √ dx = 3 x ln x Here Z Z 10 2 1 = 2 Z e 2 e eu du = eu + C = ex + C 2 2 10 ln |u| 2 ! = 1 (ln(10) − ln 2) 2 x = 0 =⇒ u = 1 + e0 = 1 + 1 = 2 2 x = ln 3 =⇒ u = 1 + e2 ln 3 = 1 + eln(3 ) = 1 + eln(9) = 1 + 9 = 10 2 5. Z π 6 0 cos x 1 dx = (1 + 6 sin x)2 6 Here 6. Z π 3 π 4 7. Z π sin 0 Here 8. Z π 8 9. u=1 1 du = cos xdx 6 Z u= 2π 3 1 du = dx 2 2 x 6 cos x= x= and x dx = 6 6 x 6
u = sin
du = 16 cos
x6 dx 6du = cos x6 dx 0 √ u = tan(2x) du = 2 sec2 (2x)dx 1 2 2 du = sec (2x)dx √ 2 x cos(x x) dx = 3 Z Z π 4 π 3 u= 12 =⇒ u = =⇒ u = 2π 3 2 π 2 3 u=0 Z 1 =− 1 1 1 3 + = = 24 6 24 8 u=1 1 2 = 0 1 4 x = 0 =⇒ u = 0 x = π =⇒ u = 12 u3 du = u=0 and 2π π 1 1 + cos = − = − cos 2 3 2 2 π 2 2π 3 u du = 2u and 1 2 4 x = 0 =⇒ u = 1 x = π6 =⇒ u = 1 + 6 sin( π6 ) = 1 + 6( 21 ) = 4 1 sin u du = − cos u 2 u= π2 u = 2x du = 2dx 1 u−2 du = − u−1 6 and tan3 (2x) sec2 (2x) dx = Here Z u=4 u = 1 + 6 sin x du = 6 cos xdx 1 sin(2x) dx = 2 Here Z 1 u4 2 4 1 = 0 1 1 −0= 8 8 x = 0 =⇒ u = 0 x = π8 =⇒ u = tan π4 = 1 cos u du = 3 2 2 sin u + C = sin(x 2 ) + C 3 3 3 Here 10. Z u = x2 1 du = 32 x 2 dx √ 2 xdx 3 du = 1 sin t2 Here   Z 1 dt = − sin u du = cos u + C = cos 1t + C t u = 1t du = − t12 dt −du = t12 dt 3  1 − 2  = 1 4 11. Z = x(x + 1) 14 dx = Z (u − 1)u 14 du = (x + 1)16 (x + 1)15 − +C 16 15 Z Here u15 − u14 du = 1 16 1 u − u15 + C 16 15 u = x + 1 =⇒ x = u − 1 du = dx Sometimes called an inverted substitution. 7 12. 7 cos(5x) − 5 sin(7x) dx = 5 7 5 sin(5x) + cos(7x) + C 5 7 Z u = 5x du = 5dx 1 5 du = dx Here 13. Z x 14. 15. 1 16. Z sin w dw = 7 5 sin u + cos w + C = 5 7 w = 7x dw = 7dx 1 7 dw = dx 1 1 1 eu du = eu + C = e− x + C 7 7 u = − x1 du = x12 dx ex dx = (ex − 1)2 Here Z Z w = 2 − 3x2 dw = −6xdx 1 − 6 dw = xdx e− x 1 dx = 2 7x 7 Here Z 5 cos u du − 7   Z p 3 3 1 3 1 √ 1 2 1 w 2 + C = − w 2 + C = − (2 − 3x2 ) 2 + C 2 − 3x2 dx = − w dw = − 6 6 3 9 9 Here Z and Z Z 1 1 du = −u−1 + C = − x +C 2 u e −1 u = ex − 1 du = ex dx 3e7x 3 √ dx = − 7x 7 1−e Here Z 1 1 3 6√ 6p √ dw = − (2)w 2 + C = − w+C = − 1 − e7x + C 7 7 7 w w = 1 − e7x dw = −7e7x dx 1 − 7 dw = e7x dx Caution: For u-substitution definite integrals, I will be changing my limits of integration (like your book does). If you choose not to change your limits of integration, then you must mark your original limits carefully. Please do not mix and match the two variables. See me with questions... 4