Exercise
Find the derivative of the following function:
f ( x ) = e x − 1 e x + 1 f(x)=\frac{e^x-1}{e^x+1} f ( x ) = e x + 1 e x − 1
Final Answer
Show final answer
f'(x)=\frac{2e^x}{{(e^x+1)}^2}
Solution
f ( x ) = e x − 1 e x + 1 f(x)=\frac{e^x-1}{e^x+1} f ( x ) = e x + 1 e x − 1
Using Derivative formulas Derivative Rules
f ′ ( x ) = e x ( e x + 1 ) − ( e x − 1 ) ⋅ e x ( e x + 1 ) 2 = f'(x)=\frac{e^x(e^x+1)-(e^x-1)\cdot e^x}{{(e^x+1)}^2}= f ′ ( x ) = ( e x + 1 ) 2 e x ( e x + 1 ) − ( e x − 1 ) ⋅ e x =
One can simplify the derivative:
= e 2 x + e x − e 2 x + e x ( e x + 1 ) 2 = =\frac{e^{2x}+e^x-e^{2x}+ e^x}{{(e^x+1)}^2}= = ( e x + 1 ) 2 e 2 x + e x − e 2 x + e x =
= 2 e x ( e x + 1 ) 2 =\frac{2e^x}{{(e^x+1)}^2} = ( e x + 1 ) 2 2 e x
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