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

Textbook Question

Chapter 3, 3.4 - Problem 15 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

Posted on

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}

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balajia | College Teacher | (Level 1) eNoter

Posted on

`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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