Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
If yx = ey-x, Prove that $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
✓
Answer
Here, yx = ey-x Taking log on both the sides, $\log\text{y}=\log\text{e}^{(\text{y}-\text{x})}$ $\big[\text{Since},\log\text{a}^{\text{b}}=\text{b}\log\text{a and}\log_\text{e}\text{e}=1\big]$ $\text{x}\log\text{y}=(\text{y}-\text{x})\log\text{e}$ $\text{x}\log\text{y}=\text{y}-\text{x}\ .....(\text{i})$ Differentiating it with respect to x using product rule, $\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\text{y}-\text{x})$ $\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=\frac{\text{dy}}{\text{dx}}-1$ $\text{x}\Big(\frac{\text{x}}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)=\frac{\text{dy}}{\text{dx}}-1$ $\frac{\text{dx}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}-1\Big)=-1-\log\text{y}$ $\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{y}}{(1+\log\text{y})\text{y}}\Big)=-(1+\log\text{y})$ $\Big[\text{Since, from equation (i), x}=\frac{\text{y}}{(1+\log\text{y})}\Big]$ $\frac{\text{dy}}{\text{dx}}\Big[\frac{1-1-\log\text{y}}{(1+\log\text{y})}\Big]=-(1+\log\text{y})$ $\frac{\text{dy}}{\text{dx}}=-\frac{(1+\log\text{y})^2}{-\log\text{y}}$ $\frac{\text{dy}}{\text{dx}}=\frac{(1+\log\text{y})^2}{\log\text{y}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.