Question :- Evaluate
[tex]\rm \: \dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{( {tan}^{ - 1} x \: + \: {cot}^{ - 1} x)} \\ \\ [/tex]
[tex]\rm \: \: \: \: \: \: (a) \: \: \: 0 \\ [/tex]
[tex]\rm \: \: \: \: \: \: (b) \: \: \: 1\\ [/tex]
[tex]\rm \: \: \: \: \: \: (c) \: \: \: e\\ [/tex]
[tex]\rm \: \: \: \: \: \: (d) \: \: \: \dfrac{\pi}{ {e}^{2} }\\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
We know,
[tex] \red{\boxed{ \rm{ \: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2} \: }}} \\ \\ [/tex]
So, using this result, we rewrite the above expression as
[tex]\rm \: = \: \dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{\dfrac{\pi}{2}} \\ \\ [/tex]
[tex] \red{\boxed{ \rm{ \: \dfrac{d}{dx} \: k \: = \: 0 \: }}} \\ \\ [/tex]
So, using this, we get
[tex]\rm \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\bf\implies \:\dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{( {tan}^{ - 1} x \: + \: {cot}^{ - 1} x)} \: = \: 0 \: \\ \\ [/tex]
Hence, optin (a) is correct.
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]
[tex] = \frac{d}{dx} ( {e}^{ {tan}^{ - 1}x + {cot}^{ - 1} x} )\\ \\ = \frac{d}{dx} {e}^{ \frac{\pi}{2} } \\ \\ = 0[/tex]
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Verified answer
Question :- Evaluate
[tex]\rm \: \dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{( {tan}^{ - 1} x \: + \: {cot}^{ - 1} x)} \\ \\ [/tex]
[tex]\rm \: \: \: \: \: \: (a) \: \: \: 0 \\ [/tex]
[tex]\rm \: \: \: \: \: \: (b) \: \: \: 1\\ [/tex]
[tex]\rm \: \: \: \: \: \: (c) \: \: \: e\\ [/tex]
[tex]\rm \: \: \: \: \: \: (d) \: \: \: \dfrac{\pi}{ {e}^{2} }\\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given expression is
[tex]\rm \: \dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{( {tan}^{ - 1} x \: + \: {cot}^{ - 1} x)} \\ \\ [/tex]
We know,
[tex] \red{\boxed{ \rm{ \: {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2} \: }}} \\ \\ [/tex]
So, using this result, we rewrite the above expression as
[tex]\rm \: = \: \dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{\dfrac{\pi}{2}} \\ \\ [/tex]
We know,
[tex] \red{\boxed{ \rm{ \: \dfrac{d}{dx} \: k \: = \: 0 \: }}} \\ \\ [/tex]
So, using this, we get
[tex]\rm \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\bf\implies \:\dfrac{d}{dx} \: {\bigg\lgroup \: e\: \bigg\rgroup }^{( {tan}^{ - 1} x \: + \: {cot}^{ - 1} x)} \: = \: 0 \: \\ \\ [/tex]
Hence, optin (a) is correct.
[tex]\rule{190pt}{2pt} \\ [/tex]
[tex] { \red{ \mathfrak{Additional\:Information}}}[/tex]
[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}[/tex]
[tex] = \frac{d}{dx} ( {e}^{ {tan}^{ - 1}x + {cot}^{ - 1} x} )\\ \\ = \frac{d}{dx} {e}^{ \frac{\pi}{2} } \\ \\ = 0[/tex]