SOLUTION: Its from differential calculus.
LEt y= 5/³√(2x-5)^2
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)
However, I can not continue on from there.
the answer at the back o
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-> SOLUTION: Its from differential calculus.
LEt y= 5/³√(2x-5)^2
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)
However, I can not continue on from there.
the answer at the back o
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Question 26736: Its from differential calculus.
LEt y= 5/³√(2x-5)^2
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)
However, I can not continue on from there.
the answer at the back on my book is:
-20
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3(2x-5)^(5/3) Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! see my working and comments on your work-----------
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Its from differential calculus.
LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N)
Y=5*(2X-5)^(-2/3)...
DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1)
DY/DX=(-20/3)*(2X-5)^(-5/3)
(-20)/3*(2X-5)^(5/3)
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So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT +
However, I can not continue on from there.
the answer at the back on my book is:
-20
----------
3(2x-5)^(5/3)
You may edit the question. Maybe convert formulae to the same formula notation as in your solutions.
