How do you use the product rule to differentiate #(2x^2-5)^3(x^3+6x)^(4/3)#?

1 Answer

Answer:

#d/dx[(2x^2-5)^3*(x^3+6x)^(4/3)]#
#=4(2x^2-5)^(2)(x^3+6x)^(1/3)(5x^4+17x^2-10)#

Explanation:

The formula #d/dx(uv)=ud/dx(v)+vd/dx(u)#

let #u=(2x^2-5)^3# and #v=(x^3+6x)^(4/3)#

Now use the formula

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))=(2x^2-5)^3d/dx((x^3+6x)^(4/3))+((x^3+6x)^(4/3))d/dx((2x^2-5)^3)#

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))=(2x^2-5)^3*(4/3)((x^3+6x)^(4/3-1))*d/dx(x^3+6x)+((x^3+6x)^(4/3))*3((2x^2-5)^(3-1))*d/dx(2x^2-5)#

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))=(2x^2-5)^3*(4/3)((x^3+6x)^(4/3-1))*(3x^2+6)+((x^3+6x)^(4/3))*3((2x^2-5)^(3-1))(4x-0)#

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))#
#=4(2x^2-5)^3*((x^3+6x)^(1/3))(x^2+2)+12x(x^3+6x)^(4/3)(2x^2-5)^(2)#

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))#
#=4(2x^2-5)^(2)(x^3+6x)^(1/3)[(2x^2-5)(x^2+2)+3x(x^3+6x)]#

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))#
#=4(2x^2-5)^(2)(x^3+6x)^(1/3)[2x^4-x^2-10+3x^4+18x^2]#

#d/dx((2x^2-5)^3*(x^3+6x)^(4/3))#
#=4(2x^2-5)^(2)(x^3+6x)^(1/3)(5x^4+17x^2-10)#

God bless....I hope the explanation is useful.