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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 21 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 21 maths textbook solution

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Answer:   \frac{d y}{d x}=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}\left[\frac{6 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2(x-3)}-\frac{2 x}{(4 x-1)}\right]

Hint:  Differentiate the equation taking log on both sides

Given:  y=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}

Solution:  

        y=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}
Taking log on both sides,

       \log y=\log \left[\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{(x-3)^{\frac{1}{2}}(4 x-1)^{\frac{1}{2}}}\right]

w.r.t x 

        \frac{1}{y} \frac{d y}{d x}=3 \log \left(x^{2}-1\right)+\log (2 x-1)-\frac{1}{2} \log (x-3)-\frac{1}{2} \log (4 x-1)

                                                                    \left[\begin{array}{l} \because \log A B=\log A+\log B \\\\ \log \frac{a}{B}=\log A-\log B \\\\ \log A^{n}=n \log A \end{array}\right]

        \frac{1}{y} \frac{d y}{d x}=\frac{3.2 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2} \cdot \frac{1}{(x-3)}-\frac{1}{2} \cdot \frac{1}{(4 x-1)} 4 x

        \frac{d y}{d x}=y\left[\frac{6 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2(x-3)}-\frac{2 x}{(4 x-1)}\right]

        \frac{d y}{d x}=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}\left[\frac{6 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2(x-3)}-\frac{2 x}{(4 x-1)}\right]

 

 

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