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If y= x/x 4 , show that. X dy/DX = y ( 1-y)?
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If y= x/x 4 , show that. X dy/DX = y ( 1-y)?
Proof:

Given, y = x/x^4

Differentiating both sides w.r.t. x,

dy/dx = d/dx (x/x^4)

Using quotient rule of differentiation,

dy/dx = [(x^4)(1) - (x)(4x^3)] / (x^4)^2

dy/dx = (x^4 - 4x^4) / x^8

dy/dx = -3x^4 / x^8

dy/dx = -3 / x^4

Multiplying both sides by x,

x dy/dx = -3/x^3

Substituting the value of y, we get

x dy/dx = -3y^2

Multiplying both sides by -1, we get

- x dy/dx = 3y^2

Rearranging, we get

x dy/dx = y(1-y)

Hence, the given expression is proved.
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If y= x/x 4 , show that. X dy/DX = y ( 1-y)?
If y=(x)/(x+4)prove that x(dy)/(dx)+y(y1)=0
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If y= x/x 4 , show that. X dy/DX = y ( 1-y)?
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