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weqwewe [10]
3 years ago
12

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Mathematics
1 answer:
Leni [432]3 years ago
8 0

Answer:  \dfrac{2x^2-1}{x(x^2-1)}

Step-by-step explanation:

The given function : y=\ln(x(x^2 - 1)^{\frac{1}{2}})

\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}    [\because \ln(ab)=\ln a +\ln b]

\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}  [\because \ln(a)^n=n\ln a]

Now , Differentiate both sides  with respect to x , we will get

\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1) (By Chain rule)

[Note : \dfrac{d}{dx}(\ln x)=\dfrac{1}{x}]

\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)

[ \because \dfrac{d}{dx}(x^n)=nx^{n-1}]

=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}

Hence, the derivative of the given function is \dfrac{2x^2-1}{x(x^2-1)} .

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