TMA 03 – MST124 Mathematics 1

Question 1

Differentiating 

(a).  f(x)=6x^2ln(4x)

f’(x)=d/dx(6x^2ln(4x))

First we remove the constant

f’(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/x

Third substitute the values into the product rule

f’(x)=6 (2xln(4x)+(1/x)*x^2)

Final Result

f’(x)=6(2xln(4x)+x) 

(b).  f(x)=

f’(x)=d/dx()

First use the quotient rule

(f/g)’=(f’*g-f*g’)/g^2

f=e^-2x-1 ,  g=cos(2x)

f’=2e^-2x , g’=-2sin(2x)

Second substitute the values in the quotient rule

f’(x)=

Final Result

f’(x)= 

(c). h(t)=

h’(t)=

First use the chain rule

dfh(u)/dt=dh/du*du/dt

h= , u=)

d/du(=1/2

d/dt())=

Second substitute the values in the chain rule

h’(t)=1/2 *)

h’(t)=1/(2)*)

Final Result

h’(t)= 

Question 2

Closed box

(a). V=L*W*H

V=2x*x*h

V=

Expression of h in terms of x if V=1000

h=1000/

(b).  A=2LW+2LH+2WH

A=2(2x*x)+2(2x*h)+2(x*h)

Final result

A=

A=

A=

A= 

(c).  A’=

A’=

0=

x=7.211248

A”=8+6000/x^3

A”=24

h=1000/2x^2

h=9.615 

Question 3 

Expansion of the expression = 

Using the Sum rule

= – 

==-cos(x)

==x

Final result

=-cos(x)-x+C 

Question 4

(a).

First using substitution

U=cos(x)

=

By taking out the constant

=

Applying substitution with V=3u-4

=

Taking the constant out

=

=

Applying the power rule

=

Substituting back all the values

V=3u-4 and u=cos(x)

=

Final result

= 

(b). 

Taking the constant out

=

Making u=2x^4 and substituting

=

=

=

We end up with

(

 =1.59726

Final Answer

1.597 

Question 5

Using integration by parts

Let u= and V=sin(x)

V’=-cos(x) , u’=2(x+3)

Hence

=

Further integrating the second part

 =

Hence

Simplifying

=

Final result

=+C 

Question 6

Area under a curve 

(a). The value of f(x) for both x’s

f(x)=

f(x)=2.021

f(x)=

f(x)=2.56

Hence, the region has positive value meaning it is above the x-axis 

(b)  Upper limit (-Lower limit (

Where the limits are the given values for upper and lower bounds 

(c).

Integration by part

Where

U=(x+1),  V=sin (5x)

U’=1,  V’=-1/5cos(5x)

Hence

=

Integrating the second part

 =

Finally

=+C

After integration

) for  =

) for  =

=

=

Area=-2.285 

Question 8

(a)

(i) AB

 (ii).   BA

Not possible, the number of columns in the first matrix must be equivalent to the number of row of the second matrix

(iii). A^2

No possible, a matrix can only be raised to a given power if the matrix has the same number of rows and columns