CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If yx + xy + xx = ab, find $\frac{\text{dy}}{\text{dx}}$
✓
Answer
Given that yx + xy + xx = ab Putting u = yx, v = xy and w = xx, we get u + v + w = ab Therefore $\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}=0\ .....(\text{i})$ Now, u = yx. Taking logrithm on both sides, we have $\log\text{u}=\text{x}\log\text{y}$ Differentiating both sides w.r.t. x, we have $\frac{1}{\text{u}}\times\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$ $=\text{x}\frac{1}{\text{y}}\times\frac{\text{dy}}{\text{dx}}+\log\text{y}\times1$ So, $\frac{\text{du}}{\text{dx}}=\text{u}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big]\ .....(\text{ii})$ Also v = xy Taking logarithm on both sides, we have $\log\text{v}=\text{y}\log\text{x}$ Differentiating both sides w.r.t. x, we have $\frac{1}{\text{v}}\times\frac{\text{dv}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}$ $=\text{y}\times\frac{1}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}$ So, $\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$ $=\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\times\frac{\text{dy}}{\text{dx}}\Big]$ Again w = xx Taking logarithm on both sides, we have $\log\text{w}=\text{x}\log\text{x}$ Differentiating both sides w.r.t x, we have $\frac{1}{\text{w}}\times\frac{\text{dw}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{x}}{\text{dx}}(\text{x})$ $=\text{x}.\frac{1}{\text{x}}+\log\text{x}\times1$ i.e. $\frac{\text{dw}}{\text{dx}}=\text{w}(1+\log\text{x})$ $=\text{x}^\text{x}(1+\log\text{x})\ .....(\text{iv})$ From (i), (ii), (iii), (iv), we have $\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)+\text{x}^\text{y}\Big(\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big) \\ +\text{x}^\text{x}(1+\log\text{x})=0$ $\big(\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\times\log\text{x}\big) \\ \frac{\text{dy}}{\text{dx}}=-\text{x}^\text{x}(1+\log\text{x})-\text{y}\times\text{x}^{\text{y}-1}-\text{y}^\text{x}\log\text{y}=0$ $\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\big[\text{y}^\text{x}\log\text{y}+\text{y}\times\text{x}^{\text{y}-1}+\text{x}^\text{x}(1+\log\text{x})\big]}{\text{x}\times\text{y}^{\text{x}-1}+\text{x}^\text{y}\log\text{y}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.