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Differentiation and integration assignment

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Added on  2021-04-24

Differentiation and integration assignment

   Added on 2021-04-24

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Question 1Differentiating(a). f(x)=6x^2ln(4x)f’(x)=d/dx(6x^2ln(4x))First we remove the constantf’(x)=6 d/dx(x^2ln(4x))Second the product rule(f*g)’=f’*g+f*g’f=x^2 , g=ln(4x)f’=d/dx(x^2)=2x , g’=d/dx(ln(4x))=1/xThird substitute the values into the product rulef’(x)=6 (2xln(4x)+(1/x)*x^2)Final Resultf’(x)=6(2xln(4x)+x)(b). f(x)=e−2x−1cos⁡(2x)f’(x)=d/dx(e−2x−1cos⁡(2x))First use the quotient rule(f/g)’=(f’*g-f*g’)/g^2f=e^-2x-1 , g=cos(2x)f’=2e^-2x , g’=-2sin(2x)Second substitute the values in the quotient rule
Differentiation and integration assignment_1
f’(x)=(−2e−2x)cos(2x)−(e−2x−1)(−2sin(2x))(cos(2x))2Final Resultf’(x)=−2e−2xcos(2x)+(2sin(2x))(e−2x−1)(cos(2x))2(c). h(t)=√(t3+sin⁡(2t))h’(t)=√(t3+sin⁡(2t))First use the chain ruledfh(u)/dt=dh/du*du/dth=√u , u=t3+sin⁡¿)d/du(√u¿=1/2√ud/dt(t3+sin⁡¿))=3t2+cos(2t)∗2Second substitute the values in the chain ruleh’(t)=1/2√u *¿)h’(t)=1/(2√t3+sin⁡(2t))*¿)Final Resulth’(t)=3t2+2cos⁡(2t)2√t3+sin⁡(2t)Question 2Closed box(a). V=L*W*HV=2x*x*hV=2x2h
Differentiation and integration assignment_2
Expression of h in terms of x if V=1000h=1000/2x2(b). A=2LW+2LH+2WHA=2(2x*x)+2(2x*h)+2(x*h)Final resultA=4x2+4xh+2xhA=4x2+4x(10002x2)+2x(10002x2)A=4x2+2000x+1000xA=4x2+3000x(c). A’=8x−3000x−2A’=8x−3000x−20=8x−3000x−2x=7.211248A”=8+6000/x^3A”=24h=1000/2x^2h=9.615Question 3∫(1−sin2x⁡sin2x)dxExpansion of the expression
Differentiation and integration assignment_3
1−sin2x⁡sin2x =1sin2x−1∫(1sin2x−1)dxUsing the Sum rule=∫1sin2xdx -∫−1dx=∫1sin2xdx=-cos(x)=∫−1dx=xFinal result=-cos(x)-x+CQuestion 4(a). ∫(sin⁡(x)(3cos(x)−4)4)dxFirst using substitutionU=cos(x)=∫−(1(3u−4)4)duBy taking out the constant=−∫(1(3u−4)4)duApplying substitution with V=3u-4=−∫(13v4)duTaking the constant out
Differentiation and integration assignment_4

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