Solutions. 1. Differentiate f(x) = (x2 +1)7. Let u = x2+ 1 Then f(x) = u2 : df du' du dx du da = 746 . 2x = 14x ( x + 1) 6 2. Differentiate - dx @ (x5 + )xp(x)+". 100 Let u = x5 + exp x and f (x) = 4100 Then of _ du" du dx du dx = 100 u". (5x4 + expx) = 100 ( x 5 + exp x ) ( 5x+ expx) easy ! "
3. Differentiate - (x5(1 + x)+2. Let U = x5 ( 1+x ) and f = (x5 (+20)2 = u2 Then of _ du' du dx du dx = 24. dx5(1+x) dx (Hx) } Lax = 2u/ dxs . (1+x) + x5 dx - = 2(x$(Hx)(5x4(1+x)+x5) d2 sin x2. 4. Compute dx2 d2f = dx2 Let u = x2 Then = d (cos u. 2x) dx = d cosu du 2x + cosu . du dx d (2x) dx = - Sin U . (2xC. 2x) + cosu . 2 = - 4x2 sin x2 + 2 cos x2 (g) = (dsinu . 44 ) du dx )
u=?? exp (u) ? Of = exp (u) . dt ???? ??????2 22 & )xp -27. 5. Compute at2 VE 6. Solve dy :14x(x2 + 1)6. dx Use the Fundamental Theorem of calculus, Szdx=[14x(x3+1)6 dx = (x2+1)+ C 4t 1 ?? 0 -1/2 72 -2 IN =? (it*+ [th) exp [- ?} (?????(S))++2ep[?] = 1 -12 - 5 z2t "2+ z't-2) exp [-] - 5 11 - 1 3t