Question

Differentiate between w.r.t.x [2x^3/2-3x^4/3-5]^5/2

Answer 1

Given: [ 2x^3/2 - 3x^4/3 - 5 ]^5/2

To find: Differentiate w.r.t. x ?

Solution:

• Now we have given the term  [ 2x^3/2 - 3x^4/3 - 5 ]^5/2. Lets consider it as y.

• So y =  [ 2x^3/2 - 3x^4/3 - 5 ]^5/2

• Differentiate it with respect to x, we get:

             dy / dx = 5/2 x [ 2x^3/2 - 3x^4/3 - 5 ]^3/2 x [3/2*2x^1/2 - 4/3*3x^1/3 ]

• Now solving further, we get:

             dy / dx = 5/2 x [ 2x^3/2 - 3x^4/3 - 5 ]^3/2 x ( 3√x - 4∛x )

             dy / dx =  [ 2x^3/2 - 3x^4/3 - 5 ]^3/2 x (15/2 √x -30∛x )

             dy / dx = [y]^3/2 x (15/2 √x -30∛x )

Answer:

            So after differentiating, we get dy / dx = [y]^3/2 x (15/2 √x -30∛x )