# `y = x e ^(-kx)` Find the derivative of the function.

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### 2 Answers

*Note*:- 1) If y = x^n ; then dy/dx = n*x6(n-1) ; where n = real number

**2) if y = e^(ax) ; then ; dy/dx = a*e^(ax)**

**3) If y = u*v ; where u & v are both functions of 'x' ; then**

**dy/dx = y' = u*(dv/dx) + v*(du/dx)**

Now,

Given y = x*e^(-kx)

thus, y' = dy/dx = -[(kx)*e^(-kx)] + {e^(-kx)}

or, dy/dx = y' = {e^(-kx)}*{1 - kx}

`y=xe^(-kx)`

`y'=e^(-kx)(dx/dx)+x(d(e^(-kx))/dx)`

`y'=e^(-kx)+x(-k)e^(-kx)`

`y'=e^(-kx)*(1-kx)`

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