How do you find the derivative of #(x^8(x-23)^(1/2))/(27x^6(4x-6)^8)#?

1 Answer
1s2s2p
Dec 8, 2017

#=(x^8(432x^5(4x-6)^8(x-23)^(1/2)+(x-23)^(-1/2)-2x^4(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7))))/(2(27x^6(4x-8)^8)^2#

Explanation:

#y=(x^8(x-23)^(1/2))/(27x^6(4x-6)^8)=f(x)/g(x)#

#dy/dx=(f'(x)g(x)-f(x)g'(x))/(f(x)^2)#

#f(x)=x^8(x-23)^(1/2)=h(x)j(x)#
#f'(x)=h'(x)j(x)+h(x)j'(x)#
#h(x)=x^8#
#h'(x)=8x^7#
#j(x)=(x-23)^(1/2)#
#j'(x)=(x-23)^(-1/2)/2#
#f'(x)=8x^7(x-23)^(1/2)+(x^8(x-23)^(-1/2))/2=(16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2=(16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2#

#g(x)=27x^6(4x-6)^8=a(x)s(x)#
#g'(x)=a'(x)s(x)+a(x)s'(x)#
#a(x)=27x^6#
#a'(x)=162x^5#
#s(x)=(4x-6)^8#
#s'(x)=4*8*(4x-6)^7=32(4x-6)^7#
#g'(x)=162x^5(4x-6)^8+27x^6 32(4x-6)^7=162x^5(4x-6)^8+864x^6(4x-6)^7=54x^5(3(4x-6)^8+16x(4x-6)^7)#

#dy/dx=(27x^6(4x-6)^8((16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2)-x^8(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7)))/(27x^6(4x-6)^8)^2#
#=(((27x^6(4x-6)^8 16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2))/2)-x^8(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7)))/(27x^6(4x-6)^8)^2#
#=(27x^6(4x-6)^8 16x^7(x-23)^(1/2)+x^8(x-23)^(-1/2)-2x^8(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7)))/(2(27x^6(4x-6)^8)^2)#
#=(x^8(432x^5(4x-6)^8(x-23)^(1/2)+(x-23)^(-1/2)-2x^4(x-23)^(1/2)(54x^5(3(4x-6)^8+16x(4x-6)^7))))/(2(27x^6(4x-8)^8)^2#

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