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  • Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 55

Need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 55

Answers (1)

Answer:   \frac{d y}{d x}=\frac{-\left(y^{x} \log y+x^{y-1} \cdot y+x^{x}(\log x+1)\right)}{\left(x^{y} \log y+y^{x-1} \cdot x\right)}

Hint: To solve this equation we solve this differently

Given: y^{x}+x^{y}+x^{x}=a^{b}

Solution:  Let uvw

        u+v+w=a^{b}

Diff w.r.t. x

        \begin{aligned} &\frac{d}{d x}(u+v+w)=\frac{d}{d x} a^{b} \\\\ &\frac{d u}{d x}+\frac{d v}{d u}+\frac{d w}{d v}=0 \\\\ &u=y^{x} \end{aligned}            ...........(1)

Taking log on both sides,

        \log u=\log y^{x} \quad\left[\because \log a^{b}=b \log a\right]

Diff w.r.t. u                              \left[\because \frac{d}{d x}(u v)=u v^{\prime}+v u^{\prime}\right]

        \frac{d}{d x}(\log u)=\frac{d}{d x}(x \log y) \quad\left[\because \frac{d}{d x} \log x=\frac{1}{x}\right]

        \begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y(1) \\\\ &\frac{d u}{d x}=u\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \end{aligned}

   

     v=x^{y}                                ..............(2)

Taking log on both sides,

        \begin{aligned} &\log v=\log x^{y} \\\\ &\log v=y \log x \end{aligned}

Diff w.r.t

        \begin{aligned} &\frac{1}{v} \frac{d v}{d x}=y \cdot \frac{1}{x}+\log x \frac{d y}{d x} \\\\ &\frac{d v}{d u}=v\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\ &w=x^{x} \end{aligned}        ..........(3)

Log on b.s.

        \begin{aligned} &\log w=\log x^{x} \\\\ &\log w=x \log x \end{aligned}

Diff w.r.t

        \frac{1}{w} \frac{d w}{d x}=x \cdot \frac{1}{x}+\log x(1)

From (1)       

          \frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0

        y^{x}\left(\frac{x}{y} d y-\log y\right)+x^{y}\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+x^{x}(1+\log x)=0

        y^{x} \log y+y^{x-1} x \frac{d y}{d x}+x^{y} \log x \frac{d y}{d x}+x^{y-1} \frac{y}{x}+x^{x}(1+\log x)=0

        y^{x} \log y+x^{y} \cdot \frac{y}{x}+x^{x}(\log x+1)+y^{x-1} x \frac{d y}{d x}+x^{y} \log x \frac{d y}{d x}=0

        y^{x-1} \frac{d y}{d x}+x^{y} \log y \frac{d y}{d x}=-\left(y^{x} \log x+x^{y} \cdot \frac{y}{x}+x^{x}(\log x+1)\right.

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

                 

 

Posted by

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