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If y=e^x e^-x then dy/dx-√y^2-4 is equal to?
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If y=e^x e^-x then dy/dx-√y^2-4 is equal to?
Solution:

Given, y = e^x e^-x

Taking logarithm on both sides,

ln y = ln (e^x e^-x)

ln y = x - x

ln y = 0

y = e^0

y = 1

Differentiating both sides with respect to x,

y = e^x e^-x

dy/dx = (d/dx)(e^x e^-x)

dy/dx = e^x (-e^-x) + e^-x (e^x)

dy/dx = 0

Finding the value of √y^2-4,

y^2 - 4 = (e^x e^-x)^2 - 4

y^2 - 4 = e^2x - 4

y^2 - 4 = e^2x - e^2x + 1-4

y^2 - 4 = -3

y^2 = 1

y = ±1

As y cannot be negative,

y = 1

Substituting the values of dy/dx and y in dy/dx-√y^2-4,

dy/dx - √y^2-4 = 0 - √(1^2-4)

dy/dx - √y^2-4 = -√3

Hence, dy/dx - √y^2-4 = -√3.
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