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Ex 11C
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Solutions
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(a)
u=3x+2v=2−x2dudx=3dvdx=−2x ddx(3x+2)(2−x2)=(3x+2)(−2x)+(2−x2)(3)=−6x2−4x+6−3x2=−9x2−4x+6
(b)
u=xv=(5x+1)7dudx=1dvdx=7(5x+1)6.(5)00000[Chain rule]=35(5x+1)6 ddx[x(5x+1)7]=(x)[35(5x+1)6]+(5x+1)7(1)=35x(5x+1)6+(5x+1)7=(5x+1)6[35x+(5x+1)]=(5x+1)6(35x+5x+1)=(5x+1)6(40x+1)
(c)
u=2x+3v=(x−3)4dudx=2dvdx=4(x−3)3.(1)00000[Chain rule]=4(x−3)3 ddx(2x+3)(x−3)4=(2x+3)[4(x−3)3]+(x−3)4(2)=(8x+12)(x−3)3+2(x−3)4=(x−3)3[(8x+12)+2(x−3)]=(x−3)3(8x+12+2x−6)=(x−3)3(10x+6)=(x−3)3(2)(5x+3)=2(x−3)3(5x+3)
(d)
u=6x2−1v=(1+5x)3dudx=6(2)xdvdx=3(1+5x)2.(5)00000[Chain rule]=12x=15(1+5x)2 ddx(6x2−1)(1+5x)3=(6x2−1)[15(1+5x)2]+(1+5x)3(12x)=(90x2−15)(1+5x)2+12x(1+5x)3=(1+5x)2[(90x2−15)+12x(1+5x)]=(1+5x)2(90x2−15+12x+60x2)=(1+5x)2(150x2+12x−15)=3(1+5x)2(50x2+4x−5)
u=(2x+3)5v=(x+2)8dudx=5(2x+3)4.(2)00000[Chain rule]dvdx=8(x+2)7.(1)00000[Chain rule]=10(2x+3)4=8(x+2)7 dydx=(2x+3)5[8(x+2)7]+(x+2)8[10(2x+3)4]=8(2x+3)5(x+2)7+10(x+2)8(2x+3)4When x=−1,dydx=8[2(−1)+3]5[(−1)+2]7+10[(−1)+2]8[2(−1)+3]4=18
u=(x+3)4v=(x−5)7dudx=4(x+3)3.(1)00000[Chain rule]dvdx=7(x−5)6.(1)00000[Chain rule]=4(x+3)3=7(x−5)6 dydx=(x+3)4[7(x−5)6]+(x−5)7[4(x+3)3]=7(x+3)4(x−5)6+4(x−5)7(x+3)3=(x+3)3(x−5)6[7(x+3)+4(x−5)]=(x+3)3(x−5)6(7x+21+4x−20)=(x+3)3(x−5)6(11x+1)When dydx=0,0=(x+3)3(x−5)6(11x+1)x=−30 or 0x=50 or 011x+1=00000000000000000−11x=−10000000000000000000.x=−111
Question 4 - Differentiate by using two different methods
Method 1: Use product rule
u=√xv=7√x−1=x12=7x12−1dudx=12x−12dvdx=7(12)x−12=12(1√x)=7(12)(1√x)=12√x=72√x ddx[√x(7√x−1)]=√x(72√x)+(7√x−1)(12√x)=7√x2√x+7√x−12√x=7√x+7√x−12√x=14√x−12√x=14√x2√x−12√x=7−12√x
Method 2: Expand, simplify and differentiate
√x(7√x−1)=7x−√x=7x−x12ddx(7x−x12)=7−12x−12=7−12(1√x)=7−12√x
(a)
u=3xv=√4−7x3=(4−7x3)12dudx=3dvdx=12(4−7x3)−12.(−21x2)=−21x22(1√4−7x3)=−21x22√4−7x3 ddx(3x√4−7x3)=(3x)(−21x22√4−7x3)+(√4−7x3)(3)=−63x32√4−7x3+3√4−7x31=−63x32√4−7x3+3√4−7x3(2√4−7x3)2√4−7x3=−63x32√4−7x3+6(4−7x3)2√4−7x3=−63x3+24−42x32√4−7x3=−105x3+242√4−7x3
(b)
u=3x−1v=√2x2+3=(2x2+3)12dudx=3dvdx=12(2x2+3)−12.(4x)=(2x)(1√2x2+3)=2x√2x2+3 ddx[(3x−1)√2x2+3]=(3x−1)(2x√2x2+3)+(√2x+3)(3)=2x(3x−1)√2x2+3+3√2x2+31=2x(3x−1)√2x2+3+3√2x2+3(√2x2+3)√2x2+3=2x(3x−1)√2x2+3+3(2x2+3)√2x2+3=2x(3x−1)+3(2x2+3)√2x2+3=6x2−2x+6x2+9√2x2+3=12x2−2x+9√2x2+3
(c)
u=2xv=√(3x−1)5=(3x−1)52dudx=2dvdx=52(3x−1)32.(3)=152(3x−1)32 ddx[2x(3x−1)52]=(2x)[152(3x−1)32]+(3x−1)52(2)=15x(3x−1)32+2(3x−1)52=(3x−1)32[15x+2(3x−1)]00000[32+1=52]=(3x−1)32(15x+6x−2)=(3x−1)32(21x−2)=√(3x−1)3(21x−2)
(d)
u=(x+2)(x−2)v=√x2+4=x2−22=(x2+4)12=x2−4dudx=2xdvdx=12(x2+4)−12.(2x)=x(1√x2+4)=x√x2+4 ddx[(x2−4)√x2+4]=(x2−4)(x√x2+4)+(√x2+4)(2x)=x(x2−4)√x2+4+2x√x2+41=x3−4x√x2+4+2x√x2+4(√x2+4)√x2+4=x3−4x√x2+4+2x(x2+4)√x2+4=x3−4x√x2+4+2x3+8x√x2+4=3x3+4x√x2+4
u=3x+2v=(2−x)−1dudx=3dvdx=(−1)(2−x)−2.(−1)=(2−x)−2=1(2−x)2 dydx=(3x+2)[1(2−x)2]+(12−x)(3)=3x+2(2−x)2+32−x=3x+2(2−x)2+3(2−x)(2−x)2=3x+2+3(2−x)(2−x)2=3x+2+6−3x(2−x)2=8(2−x)2When dydx=8,8=8(2−x)28(2−x)2=8(2−x)2=88(2−x)2=12−x=±√12−x=±1 2−x=1 or 2−x=−1−x=−1−x=−3x=1x=3
Since the y-axis is a vertical line, the tangent to the curve must be a horizontal line (with gradient = 0)
u=2x−3v=(x+5)5dudx=2dvdx=5(x+5)4.(1)=5(x+5)4 dydx=(2x−3)[5(x+5)4]+(x+5)5(2)=5(2x−3)(x+5)4+2(x+5)5=(x+5)4[5(2x−3)+2(x+5)]=(x+5)4(10x−15+2x+10)=(x+5)4(12x−5)When dydx=0,00000[Gradient of tangent = 0]0=(x+5)4(12x−5) (x+5)4=0 or 12x−5=0x+5=012x=5x=−5x=512
u=x+av=√b−x=(b−x)12dudx=1dvdx=12(b−x)−12.(−1)=−12(1√b−x)=−12√b−x dydx=(x+a)(−12√b−x)+(√b−x)(1)=−(x+a)2√b−x+√b−x1=−x−a2√b−x+√b−x(2√b−x)2√b−x=−x−a2√b−x+2(b−x)2√b−x=−x−a2√b−x+2b−2x2√b−x=−3x−a+2b2√b−xSubstitute x=14 and dydx=0,0=−3(14)−a+2b2√b−(14)0=−34−a+2bLet a=14,0=−34−14+2b−2b=−1b=−1−2=12Possible values: a=14,b=12
(i) Note we cannot take the square root of a negative number. For example, √−1 has no real solution.
For √9−2x2 to have real roots,9−2x2≥0−2x2≥−9x2≤92x2≤(3√2)2∴−3√2≤.x≤3√2
(ii)
u=x2v=√9−2x2=(9−2x2)12dudx=2xdvdx=12(9−2x2)−12.(−4x)=−2x(1√9−2x2)=−2x√9−2x2 dydx=(x2)(−2x√9−2x2)+(√9−x2)(2x)=−2x3√9−2x2+2x√9−x21=−2x3√9−2x2+2x√9−2x2(√9−2x2)√9−2x2=−2x3√9−2x2+2x(9−2x2)√9−2x2=−2x3√9−2x2+18x−4x3√9−2x2=−6x3+18x√9−2x2To show ydydx+6x3(x2−3)=0,L.H.S=(x2√9−2x2)(−6x3+18x√9−2x2)+6x3(x2−3)=x2(−6x3+18x)+6x5−18x3=−6x5+18x3+6x5−18x3=0=R.H.S