Exercise
Find the derivative of the function
f ( x ) = ln ( x − 1 x + 1 ) f(x)=\ln (\sqrt{\frac{x-1}{x+1}}) f ( x ) = ln ( x + 1 x − 1 )
Final Answer
Show final answer
f'(x)=\frac{1}{x^2-1}
Solution
f ( x ) = ln ( x − 1 x + 1 ) f(x)=\ln (\sqrt{\frac{x-1}{x+1}}) f ( x ) = ln ( x + 1 x − 1 )
Using derivative formulas Derivative Rules
f ′ ( x ) = 1 x − 1 x + 1 ⋅ 1 2 x − 1 x + 1 ⋅ 1 ⋅ ( x + 1 ) − ( x − 1 ) ⋅ 1 ( x + 1 ) 2 = f'(x)=\frac{1}{\sqrt{\frac{x-1}{x+1}}}\cdot\frac{1}{2\sqrt{\frac{x-1}{x+1}}}\cdot\frac{1\cdot (x+1)-(x-1)\cdot 1}{{(x+1)}^2}= f ′ ( x ) = x + 1 x − 1 1 ⋅ 2 x + 1 x − 1 1 ⋅ ( x + 1 ) 2 1 ⋅ ( x + 1 ) − ( x − 1 ) ⋅ 1 =
One can simplify the derivative:
= 1 2 ⋅ ( x − 1 x + 1 ) ⋅ x + 1 − x + 1 ( x + 1 ) 2 = =\frac{1}{2\cdot (\frac{x-1}{x+1})}\cdot\frac{x+1-x+1}{{(x+1)}^2}= = 2 ⋅ ( x + 1 x − 1 ) 1 ⋅ ( x + 1 ) 2 x + 1 − x + 1 =
= 1 2 ⋅ ( x − 1 x + 1 ) ⋅ 2 ( x + 1 ) 2 = =\frac{1}{2\cdot (\frac{x-1}{x+1})}\cdot\frac{2}{{(x+1)}^2}= = 2 ⋅ ( x + 1 x − 1 ) 1 ⋅ ( x + 1 ) 2 2 =
= 1 x − 1 x + 1 ⋅ 1 ( x + 1 ) 2 = =\frac{1}{\frac{x-1}{x+1}}\cdot\frac{1}{{(x+1)}^2}= = x + 1 x − 1 1 ⋅ ( x + 1 ) 2 1 =
= x + 1 x − 1 ⋅ 1 ( x + 1 ) 2 = =\frac{x+1}{x-1}\cdot\frac{1}{{(x+1)}^2}= = x − 1 x + 1 ⋅ ( x + 1 ) 2 1 =
= 1 x − 1 ⋅ 1 x + 1 = =\frac{1}{x-1}\cdot\frac{1}{x+1}= = x − 1 1 ⋅ x + 1 1 =
= 1 ( x − 1 ) ( x + 1 ) = =\frac{1}{(x-1)(x+1)}= = ( x − 1 ) ( x + 1 ) 1 =
Using a short multiplication formula
= 1 x 2 − 1 =\frac{1}{x^2-1} = x 2 − 1 1
Have a question? Found a mistake? – Write a comment below!here
