Home > Calculus > Quotient Rule

How do you find the derivative of #[e^x / (1 - e^x)]#?

Answer 1

Charles Anctil

#y'=e^x/(1-e^x)^2#.

Let #y=e^x/(1-e^x)#.

To find #y'#, we will use the Quotient Rule , which states, :

#(u/v)'=(vu'-uv')/v^2#.

#y=e^x/(1-e^x)#

#rArr y'={(1-e^x)(e^x)'-e^x(1-e^x)'}/(1-e^x)^2#.

#={(1-e^x)(e^x)-e^x(1'-(e^x)')}/(1-e^x)^2#.

#={(1-e^x)e^x-e^x(0-e^x)}/(1-e^x)^2#.

#=(e^x-e^(2x)+e^(2x))/(1-e^x)^2#

#:. y'=e^x/(1-e^x)^2#.

Enjoy Maths.!

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Scarlett Allison

To find the derivative of ( \frac{e^x}{1 - e^x} ), apply the quotient rule. Let ( u = e^x ) and ( v = 1 - e^x ). Then differentiate both ( u ) and ( v ) with respect to ( x ), substitute into the quotient rule formula, and simplify the result.

The derivative of ( \frac{e^x}{1 - e^x} ) is given by:

[ \frac{d}{dx} \left( \frac{e^x}{1 - e^x} \right) = \frac{e^x(1 - e^x) - e^x(-e^x)}{(1 - e^x)^2} ]

Simplify the expression:

[ \frac{e^x - e^{2x} + e^{2x}}{(1 - e^x)^2} ]

[ \frac{e^x}{(1 - e^x)^2} ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.