Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,433 questions

17,804 answers

1,438 comments

47,216 users

How to differentiate these functions?

+1 vote

1. 1/(4x - 2)^3

2. 6/√(3x - 5)

asked Feb 13, 2013 in CALCULUS by potatoes Rookie

3 Answers

0 votes
 
Best answer

1) f(x)=1/(4x - 2)^3

Using the Quotient Rule of differention,

d/dx(u/v)=(vu'-uv')/v^2

f'(x)=((4x - 2)^3)*(d/dx(1))-(1*(d/dx(4x - 2)^3)/((4x - 2)^3)^2

         =0-(3(4x - 2)^2*(4))/((4x - 2)^6)

         =-12((4x - 2)^2/((4x - 2)^6

         =-12/(4x - 2)^4

Differention of function is -12/(4x - 2)^4

answered Feb 14, 2013 by bradely Mentor
selected Feb 14, 2013 by potatoes
+3 votes

2) f(x)=6/√(3x - 5)

Using the Quotient Rule of differention, d/dx(u/v)=(vu'-uv')/v^2

f'(x)=((√(3x - 5))*(d/dx(6))-(1*(d/dx(√(3x - 5))/((√(3x - 5))^2

      =0-((1/2)(3x - 5)^(-1/2)*(3))/3x - 5

     =-(3/2)((3x - 5)^(-1/2))/3x - 5

       =-(3/2)(3x - 5)^(-3/2)

Differentiation of function is -(3/2)(3x - 5)^(-3/2)

I hope that helps u

answered Feb 14, 2013 by bradely Mentor

Differentiation of 6/√(3x - 5) is - 9/(3x - 5)3/2 .

0 votes
  • 1). f(x) = 1/(4x - 2)3 .

f(x) = (4x - 2)- 3

Power rule for derivatives : if f(x) = xn, then f '(x) = nxn - 1.

f '(x) = - 3(4x - 2)- 3 - 1 (4x - 2) '

= - 3(4x - 2)- 4 (4)

= - 12/(4x - 2)4 .

 ∴ Differentiation of 1/(4x - 2)3 is -12/(4x - 2)4 .

  • 2). f(x) = 6/√(3x - 5).

f(x) = 6(3x - 5)-1/2

Power rule for derivatives : if f(x) = xn, then f '(x) = nxn - 1.

f '(x) = 6[(-1/2)(3x - 5)-1/2 - 1]*(3x - 5) '

= - 3 * (3x - 5)-3/2 * 3

= - 9/(3x - 5)3/2 .

 ∴ Differentiation of 6/√(3x - 5) is - 9/(3x - 5)3/2 .

answered Jul 8, 2014 by lilly Expert

Related questions

asked May 17, 2014 in CALCULUS by anonymous
asked Mar 4, 2015 in CALCULUS by anonymous
asked Oct 11, 2014 in CALCULUS by anonymous
asked Oct 4, 2014 in CALCULUS by anonymous
asked Jul 22, 2014 in CALCULUS by anonymous
...