Answer:
This is done by chain rule.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \: y \: = \: {e}^{(1 + log_{e}x )} \\ [/tex]
We know,
[tex]\boxed{ \rm{ \: log_{e}(e) = 1 \: }} \\ \\ [/tex]
So, using this, result we get
[tex]\rm \: y \: = \: {e}^{( log_{e}e + log_{e}x )} \\ [/tex]
[tex]\boxed{ \rm{ \:logx + logy = log(xy) \: }} \\ \\ [/tex]
So, using this result, we get
[tex]\rm \: y \: = \: {e}^{log_{e}ex} \\ [/tex]
[tex]\boxed{ \rm{ \: {e}^{ log_{e}(x) } = x \: }} \\ \\ [/tex]
[tex]\rm \: y = ex \\ [/tex]
On differentiating both sides w. r. t. x, we get
[tex]\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}(ex) \\ [/tex]
[tex]\rm \: \dfrac{dy}{dx} = e\dfrac{d}{dx}x \\ [/tex]
[tex]\rm \: \dfrac{dy}{dx} = e \times 1 \\ [/tex]
[tex]\rm\implies \:\bf \: \dfrac{dy}{dx} = e\\ \\ [/tex]
[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]
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Answers & Comments
Answer:
This is done by chain rule.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \: y \: = \: {e}^{(1 + log_{e}x )} \\ [/tex]
We know,
[tex]\boxed{ \rm{ \: log_{e}(e) = 1 \: }} \\ \\ [/tex]
So, using this, result we get
[tex]\rm \: y \: = \: {e}^{( log_{e}e + log_{e}x )} \\ [/tex]
We know,
[tex]\boxed{ \rm{ \:logx + logy = log(xy) \: }} \\ \\ [/tex]
So, using this result, we get
[tex]\rm \: y \: = \: {e}^{log_{e}ex} \\ [/tex]
We know,
[tex]\boxed{ \rm{ \: {e}^{ log_{e}(x) } = x \: }} \\ \\ [/tex]
So, using this result, we get
[tex]\rm \: y = ex \\ [/tex]
On differentiating both sides w. r. t. x, we get
[tex]\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}(ex) \\ [/tex]
[tex]\rm \: \dfrac{dy}{dx} = e\dfrac{d}{dx}x \\ [/tex]
[tex]\rm \: \dfrac{dy}{dx} = e \times 1 \\ [/tex]
[tex]\rm\implies \:\bf \: \dfrac{dy}{dx} = e\\ \\ [/tex]
[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]