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Question Number 150881 by mathdanisur last updated on 16/Aug/21

$$\mathrm{y}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathbf{x}}\cdot {\left(\frac{\mathrm{b}}{\mathrm{x}}\right)}^{\mathbf{a}}\cdot {\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\mathbf{b}}\Rightarrow \mathrm{y}{}^{{}^{\prime }}=?$$

Answered by john_santu last updated on 16/Aug/21

$$\mathrm{y}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{x}}{\left(\frac{\mathrm{x}}{\mathrm{b}}\right)}^{\mathrm{b}-\mathrm{a}}$$$$\mathrm{let}\{\begin{array}{l}\mathrm{u}={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{x}}\Rightarrow \ln \mathrm{u}=\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\ln \mathrm{x}\\ \mathrm{v}={\left(\frac{\mathrm{x}}{\mathrm{b}}\right)}^{\mathrm{b}-\mathrm{a}}\end{array}$$$${\mathrm{y}}^{\prime }={\mathrm{u}}^{\prime }\mathrm{v}+\mathrm{u}{\mathrm{v}}^{\prime }$$$$\frac{{\mathrm{u}}^{\prime }}{\mathrm{u}}=\frac{\mathrm{a}}{\mathrm{bx}}\Rightarrow {\mathrm{u}}^{\prime }={\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{x}}\left(\frac{\mathrm{a}}{\mathrm{bx}}\right)$$$${\mathrm{v}}^{\prime }=(\mathrm{b}-\mathrm{a})\left(\frac{1}{\mathrm{b}}\right){\left(\frac{\mathrm{x}}{\mathrm{b}}\right)}^{\mathrm{b}-\mathrm{a}-1}$$$$$$

Commented by mathdanisur last updated on 16/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser},\mathrm{then}.?$$

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