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Question Number 81565 by jagoll last updated on 14/Feb/20

$$Iff\left(x\right)={\left(\tan x\right)}^{\cot x}+{\left(\cot x\right)}^{\tan x}$$$$f{}^{\prime }\left(x\right)=?$$

Commented by john santu last updated on 14/Feb/20

$$\mathrm{good}\mathrm{question}$$

Commented by mathmax by abdo last updated on 14/Feb/20

$$f\left(x\right)={\left(tanx\right)}^{cotanx}+{\left(cotanx\right)}^{tanx}\Rightarrow$$$$f\left(x\right)=e^{\frac{1}{tanx}ln\left(tanx\right)}+e^{tanxln\left(\frac{1}{tanx}\right)}=e^{\frac{ln\left(tanx\right)}{tanx}}+e^{-tanxln\left(tanx\right)}$$$$=u\left(x\right)+v\left(x\right)\Rightarrow f^{{}^{\prime }}\left(x\right)=u^{{}^{\prime }}\left(x\right)+v^{{}^{\prime }}\left(x\right)$$$$u^{{}^{\prime }}\left(x\right)={\left(\frac{ln\left(tanx\right)}{tanx}\right)}^{{}^{\prime }}u\left(x\right)=\{\frac{(1+{tan}^{2}x)}{tanx}\times tanx-ln\left(tanx\right)(1+{tan}^{2}x)\}\times \frac{1}{{tan}^{2}x}\times u\left(x\right)$$$$=\{1+{tan}^{2}x-ln\left(tanx\right)-ln\left(tanx\right){tan}^{2}x\}\times \frac{u\left(x\right)}{{tan}^{2}x}$$$$v^{{}^{\prime }}\left(x\right)={(-tanxln\left(tanx\right))}^{{}^{\prime }}v\left(x\right)$$$$=-\{(1+{tan}^{2}x)ln\left(tanx\right)+tanx\times \frac{1+{tan}^{2}x}{tanx}\}v\left(x\right)$$$$=-\{(1+{tan}^{2}x)ln\left(tanx\right)+1+{tan}^{2}x\}v\left(x\right)$$$$$$

Answered by john santu last updated on 14/Feb/20

Commented by jagoll last updated on 14/Feb/20

$$thankalotof$$

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