Question-119565

Question Number 119565 by bemath last updated on 25/Oct/20

Commented by bemath last updated on 25/Oct/20

$${gave}\:{kudos}\: \\ $$

Answered by TANMAY PANACEA last updated on 25/Oct/20

$${y}={log}_{\mathrm{7}} \left(\frac{{x}+\mathrm{6}}{{x}+\mathrm{6}}\right)^{{ln}\mathrm{7}} \\ $$$${y}={ln}\mathrm{7}×\frac{{ln}\left(\frac{{x}+\mathrm{6}}{{x}−\mathrm{6}\:}\right)}{{ln}\mathrm{7}}\:\:\:\:\:\:\:\:\:\left[{log}_{{a}} {b}=\frac{{logb}}{{loga}}\right) \\ $$$${y}={ln}\left({x}+\mathrm{6}\right)−{ln}\left({x}−\mathrm{6}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}+\mathrm{6}}−\frac{\mathrm{1}}{{x}−\mathrm{6}}=\frac{{x}−\mathrm{6}−{x}−\mathrm{6}}{\left({x}+\mathrm{6}\right)\left({x}−\mathrm{6}\right)} \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{12}}{\left({x}+\mathrm{6}\right)\left({x}−\mathrm{6}\right)} \\ $$

Answered by Ar Brandon last updated on 25/Oct/20

$$\mathrm{y}=\mathrm{log}_{\mathrm{7}} \left[\left(\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}−\mathrm{6}}\right)^{\mathrm{ln7}} \right] \\ $$$$\mathrm{7}^{\mathrm{y}} =\left(\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}−\mathrm{6}}\right)^{\mathrm{ln7}} =\left(\mathrm{1}+\frac{\mathrm{12}}{\mathrm{x}−\mathrm{6}}\right)^{\mathrm{ln7}} \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{y}} \mathrm{ln7}\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{ln7}\frac{\mathrm{12}}{\left(\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} }\left(\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}−\mathrm{6}}\right)^{\mathrm{ln7}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=−\frac{\mathrm{12}}{\left(\mathrm{x}−\mathrm{6}\right)\left(\mathrm{x}+\mathrm{6}\right)}\left(\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}−\mathrm{6}}\right)^{\mathrm{ln7}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{−\mathrm{12}}{\left(\mathrm{x}−\mathrm{6}\right)\left(\mathrm{x}+\mathrm{6}\right)} \\ $$

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