# Differentiation and integration assignment

11 Pages783 Words60 Views

## Differentiation and integration assignment

BookmarkShareRelated Documents
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)=e2x1cos(2x)f’(x)=d/dx(e2x1cos(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
f’(x)=(2e2x)cos(2x)(e2x1)(2sin(2x))(cos(2x))2Final Resultf’(x)=2e2xcos(2x)+(2sin(2x))(e2x1)(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¿=1/2ud/dt(t3+sin¿))=3t2+cos(2t)2Second substitute the values in the chain ruleh’(t)=1/2u *¿)h’(t)=1/(2t3+sin(2t))*¿)Final Resulth’(t)=3t2+2cos(2t)2t3+sin(2t)Question 2Closed box(a). V=L*W*HV=2x*x*hV=2x2h
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’=8x3000x2A’=8x3000x20=8x3000x2x=7.211248A”=8+6000/x^3A”=24h=1000/2x^2h=9.615Question 3(1sin2xsin2x)dxExpansion of the expression
1sin2xsin2x =1sin2x1(1sin2x1)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(3u4)4)duBy taking out the constant=(1(3u4)4)duApplying substitution with V=3u-4=(13v4)duTaking the constant out

## End of preview

Want to access all the pages? Upload your documents or become a member.

Related Documents
|11
|1417
|131

|5
|1672
|442

|10
|935
|80

|24
|1791
|310

|10
|2101
|439

|11
|1422
|23

### Support

#### +1 306 205-2269

Chat with our experts. we are online and ready to help.