How do you differentiate # y = 4(x-(3-5x)^4)^2# using the chain rule?

1 Answer

Answer:

#dy/dx=8x+160x(3-5x)^3-8(3-5x)^4-160(3-5x)^7#

Explanation:

let us start from the given equation #y=4(x-(3-5x)^4)^2#

#dy/dx=d/dx4(x-(3-5x)^4)^2=4*d/dx(x-(3-5x)^4)^2#

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

#dy/dx=8*(x-(3-5x)^4)^(1)*[d/dx(x)-d/dx(3-5x)^4]#

#dy/dx=8*(x-(3-5x)^4)*[1-4(3-5x)^(4-1)*d/dx(3-5x)]#

#dy/dx=8*(x-(3-5x)^4)*[1-4(3-5x)^(3)(0-5)]#

#dy/dx=8*(x-(3-5x)^4)*[1+20(3-5x)^(3)]#

#dy/dx=8(x-(3-5x)^4+20x(3-5x)^3-20(3-5x)^7)#

#dy/dx=8x+160x(3-5x)^3-8(3-5x)^4-160(3-5x)^7#

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