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  • Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 61 maths textbook solution

Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 61 maths textbook solution

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Answer: \frac{d y}{d x} \frac{\alpha}{(1-\alpha x)^{2}}+\frac{\beta\left(1-\beta x^{2}\right)}{(1-x)^{2}\left(1-\beta x^{2}\right)}+\frac{2 \alpha \beta y^{2} x^{3}-\alpha \beta y x-\alpha y^{2} x^{2}-y^{3} \beta x^{3}}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-y x)^{2}}

Hint: To solve this equation we differentiate it differently

Given: y=1+\frac{\alpha}{\left(\frac{1}{x}-\alpha\right)}+\frac{\frac{\beta}{x}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)}+\frac{\frac{\gamma}{x^{2}}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)\left(\frac{1}{x}-\gamma\right)}

Solution:  

        y=1+\frac{\alpha}{\left(\frac{1}{x}-\alpha\right)}+\frac{\frac{\beta}{x}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)}+\frac{\frac{\gamma}{x^{2}}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)\left(\frac{1}{x}-\gamma\right)}

        y=1+A+B+C

        \frac{d y}{d x}=0+\frac{d A}{d x}+\frac{d B}{d x}+\frac{d C}{d x}

        \begin{aligned} &A=\frac{\alpha x}{(1-\alpha x)} \\\\ &\frac{d A}{d x}=\frac{(1-\alpha x) \alpha-\alpha x(-\alpha)}{(1-\alpha x)^{2}} \end{aligned}

        \begin{aligned} &\frac{d A}{d x}=\frac{\alpha-\alpha^{2} x+\alpha^{2} x}{(1-\alpha x)^{2}} \\\\ &\frac{d A}{d x}=\frac{\alpha}{(1-\alpha x)^{2}} \end{aligned}        .............(1)

        \frac{d B}{d x}=\frac{d}{d x}\left(\frac{\beta x}{(1-x)(1-\beta x)}\right)

        \frac{d B}{d x}=\frac{(1-x)(1-\beta x) \beta-\beta x(1-x)(-\beta)+(1-\beta x)(-1)}{(1-x)^{2}(1-\beta x)^{2}}

        \frac{d B}{d x}=\frac{\beta\left(1-x-\beta x+\beta x^{2}\right)-\beta x(-\beta+\beta x-1+\beta x)}{(1-x)^{2}(1-\beta x)^{2}}

        \frac{d B}{d x}=\frac{\left.\beta-\beta x-\beta^{2} x+\beta^{2} x^{2}+\beta^{2} x-\beta^{2} x^{2}+\beta x-\beta^{2} x^{2}\right)}{(1-x)^{2}(1-\beta x)^{2}}

        \frac{d B}{d x}=\frac{\beta\left(1-\beta x^{2}\right)}{(1-x)^{2}(1-\beta x)^{2}}                .................(2)

  c=\frac{y x}{(1-\alpha x)(1-\beta x)(1-\gamma x)}

  \frac{d c}{d x}=\frac{(1-\alpha x)(1-\beta x)(1-\gamma x) y-y x(1-\alpha x)(1-\beta x)-\gamma(1-\beta)(1-\gamma x)(-\alpha)+(1-\alpha x)(1-\gamma x)(-\beta)}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-\gamma x)^{2}}

  \frac{dc}{dx}=\frac{\begin{aligned} &\left(1-\alpha x-\beta x+\alpha \beta x^{2}\right)(1-\gamma x) \gamma-\gamma x\{(-\gamma)(1-\alpha x-\beta x+\alpha \beta x)+(-\alpha)+(1-\beta x-\gamma x+\beta \gamma x)+ &\left.(-\beta)\left(1-\alpha x-\gamma x+\alpha \gamma x^{2}\right)\right\} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}

        

  \frac{dc}{dx}=\frac{\begin{aligned} &\left(1-\alpha x-\beta x+\alpha \beta x^{2}\right)(1-\gamma x) \gamma- &\gamma x\left\{(-\gamma)(1-\alpha x-\beta x+\alpha \beta x)+(-\alpha)+\left(1-\beta x-\gamma x+\beta \gamma x^{2}\right)+\right. &\left.(-\beta)\left(1-\alpha x-\gamma x+\alpha \gamma x^{2}\right)\right\} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}

 \frac{dc}{dx}=\frac{\begin{aligned} &\gamma\left(1-\alpha x-\beta x+\alpha \beta x^{2}-\gamma x+\alpha y x^{2}+\beta \gamma x^{2}-\alpha \beta \gamma x^{2}\right. &-\gamma x\left(-\gamma+\alpha \gamma x+\beta \gamma x-\alpha \beta \gamma^{2} x^{2}-\alpha+\alpha \beta x+\alpha \gamma x-\alpha \beta y x^{2}-\right. &\left.\left.\beta+\alpha \beta x+\gamma \beta x+\alpha \beta \gamma x^{2}\right)\right\} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}

\frac{dc}{dx}=\frac{\begin{aligned} &\gamma-\alpha \gamma x-\beta \gamma x+\alpha \beta \gamma x^{2}-\gamma^{2} x+\alpha \gamma^{2} x^{2}+\beta \gamma^{2} x^{2} &-\alpha \beta \gamma^{2} x^{2}-\gamma^{2} x-\alpha \gamma^{2} x^{2}-\beta \gamma^{2} x^{2}+\alpha \beta \gamma^{2} x^{3}+ &\alpha \gamma x-\alpha \beta \gamma x-\alpha \gamma^{2} x^{2}+\alpha \beta y^{2} x^{3}+\beta \gamma x- &\alpha \beta \gamma x^{2}+\gamma^{2} \beta x^{2}+\alpha \beta \gamma^{2} x^{3} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}

\frac{d c}{d x}=\frac{-\alpha \beta \gamma x-\alpha \gamma^{2} x^{2}-\alpha \beta \gamma^{2} x^{3}-\beta \gamma^{2} x^{2}}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-\gamma x)^{2}}

\frac{d y}{d x}=\frac{\alpha}{(1-\alpha x)^{2}}+\frac{\beta\left(1-\beta x^{2}\right)}{(1-x)^{2}(1-\beta x)^{2}}+\frac{+2 \alpha \beta \gamma^{2} x^{3}-\alpha \beta \gamma x-\alpha \gamma^{2} x^{2}-\beta \gamma^{3} x^{3}}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-\gamma x)^{2}}

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