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I am trying to implicitly differentiate the following function: $$ \lambda = \exp \left[ \left( \alpha + \frac{s}{\lambda-s} \right)^{-1} \right] $$ Can someone help me with this?

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  • $\begingroup$ Mathematica gives me $\dfrac{\mathrm d\lambda}{\mathrm d\alpha} = \dfrac{(s-\lambda )^2}{s-\exp[-(\alpha +s/(\lambda -s))^{-1}] (\alpha (\lambda - s)+s)^2}$. $\endgroup$
    – user856
    Jul 6, 2015 at 19:45

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Before making the differentiation you may simplify the equation such as

$$\ln(\lambda)=-\bigg(\alpha+\frac{s}{\lambda-s}\bigg)$$ $$\ln(\lambda)=-\bigg(\frac{\alpha(\lambda-s)+ s}{\lambda-s}\bigg)$$ $$(s-\lambda )\ln(\lambda)=\alpha(\lambda-s)+ s$$

Now you can differentiate as below

$$-\ln(\lambda)d\lambda+\frac{s-\lambda }{\lambda}d\lambda=(\lambda-s)d\alpha+\alpha d\lambda$$ $$\bigg(\alpha+\frac{s-\lambda }{\lambda}-\ln(\lambda)\bigg)d\lambda=(\lambda-s)d\alpha$$ $$\rightarrow \frac{d\lambda}{d\alpha}=\frac{(\lambda-s)}{\alpha+\frac{s-\lambda }{\lambda}-\ln(\lambda)}$$

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