Derivatives involving ln x Find the following derivatives. ddx(ln⁡(x+1x−1))

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Derivatives involving ln x Find the following derivatives.  ddx(ln⁡(x+1x−1))

boronganfh · Accepted Answer

Step 1  Given:  ddx(ln⁡(x+1x−1))  To Find: To evaluate the given derivative.  Step 2  Solution:  Let y=ln⁡(x+1x−1)  We know that ln⁡(mn)=ln⁡m−ln⁡n   (Properties of  logarithms)  Hence y=ln⁡(x+1)−ln⁡(x−1)  Differentiating with respect to x we get  dydx=ddx[ln⁡(x+1)−ln⁡(x−1)]  dydx=ddx[ln⁡(x+1)]−ddx[ln⁡(x−1)]  dydx=1x+1−1x−1  dydx=(x−1)−(x+1)(x+1)(x−1)  dydx=x−1−x−1(x+1)(x−1)  dydx=−2x2−1   [Using (a+b)(a−b)=a2−b2]  Hence ddx(ln⁡(x+1x−1))=−2x2−1

Orlando Paz · Accepted Answer

The function can also be written as:f(x)=ln⁡(x+1x−1)=ln⁡(x+1)−ln⁡(x−1)so we have:ddx(ln⁡(x+1)−ln⁡(x−1))=(ln⁡(x+1)−ln⁡(x−1))′=ln⁡(x+1)′−(ln⁡(x−1))′=1x+1⋅1−1x−1⋅1=(x−1)⋅1−(x+1)⋅1(x+1)(x−1)=x−1−x−1(x+1)(x−1)=−2x2−1Thus,ddx(ln⁡(x+1x−1))=−2x2−1Result:−2x2−1

alenahelenash · Accepted Answer

Solution:Since the given function represents a composition of two functions we must use the Chain Rule to obtain the derivative.If we denote f(x)=ln⁡x and g(x)=x+1x−1 we will have that:(f∗g)(x)=f(g(x))=f(x+1x−1)=ln⁡(x+1x−1)ddx(ln⁡(x+1x−1))=ddx(f(g(x)))=f&#039;(g(x))*g&#039;(x) (Chain Rule)=1x+1x−1∗(x+1x−1)′=x−1x+1∗(x+1)′(x−1)−(x+1)(x−1)′(x−1)2 (Quotient Rule)=x−1x+1∗(x′+1′)(x−1)−(x+1)(x′−1′)(x−1)2 (Sum Rule)=1x+1∗1∗(x−1)−(x+1)∗1x−1=1x+1∗x−1−x−1x−1=−2x2−1Result:ddx(ln⁡(x+1x−1))=21−x2

Derivatives involving ln x Find the following derivatives. \frac{d}{dx}(\ln (\frac{x+1}{x-1}))

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