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Sagot :
Answer:
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-}}[/tex]
☼︎~Given function is
[tex]\rm \: f(x) = log(1 + x) - \dfrac{2x}{x + 2}[/tex]
~Let first define the domain of f(x).
[tex]{ \underline { \underline{ {\rule{200cm}{0.4cm}}}}}[/tex]
- Now, log(1 + x) is defined when x + 1 > 0
☼︎~Now, Consider
[tex]\sf \: f(x) = log(1 + x) - \dfrac{2x}{x + 2}[/tex]
☼︎~On differentiating both sides
w. r. t. x, we get
[tex]\rm {\: \longmapsto\dfrac{d}{dx}f(x) =\dfrac{d}{dx}\bigg[ log(1 + x) - \dfrac{2x}{x + 2}}[/tex]
☼︎~We know,
[tex]\boxed{\sf{ \longmapsto\dfrac{d}{dx}logx \: = \: \dfrac{1}{x}}}[/tex]
❀And
[tex]\boxed{\sf{\longmapsto \dfrac{d}{dx} \dfrac{u}{v} = \dfrac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }}[/tex]
~So, using these results, we get
[tex]\sf {\:\longmapsto f'(x) = \dfrac{1}{x + 1} - \dfrac{(x + 2)\dfrac{d}{dx}2x - 2x\dfrac{d}{dx}(x + 2)}{ {(x + 2)}^{2} }}[/tex]
☼︎~We know,
[tex]\boxed{\rm{\longmapsto \dfrac{d}{dx}x = \sf \pink{ 1 \:} }} [/tex]
And-
[tex]\begin{gathered}\boxed{\sf{\longmapsto\dfrac{d}{dx}k = 0 \: }} \\ \end{gathered}[/tex]
☼︎~So, using this result, we get
[tex]\rm \: f'(x) = \dfrac{1}{x + 1} - \dfrac{(x + 2)2 - 2x(1 + 0)}{ {(x + 2)}^{2} }f′(x)=x+11−(x+2)2(x+2)2−2x(1+0)
[tex]\rm \: f'(x) = \dfrac{1}{x + 1} - \dfrac{(x + 2)2 - 2x(1 + 0)}{ {(x + 2)}^{2} }f′(x)=x+11−(x+2)2(x+2)2−2x(1+0)\rm \: f'(x) = \dfrac{1}{x + 1} - \dfrac{2x + 4 - 2x}{ {(x + 2)}^{2} }[/tex]
[tex]\rm \: f'(x) = \dfrac{1}{x + 1} - \dfrac{4 }{ {(x + 2)}^{2} }f′(x)=x+11−(x+2)24[/tex]
[tex]\rm \: f'(x) = \dfrac{ {(x + 2)}^{2} - 4(x + 1)}{(x + 1) {(x + 2)}^{2} }[/tex]
[tex]\rm \: f'(x) = \dfrac{ {x}^{2} + 4 + 4x - 4x - 4}{(x + 1) {(x + 2)}^{2} }[/tex]
[tex]\rm \: f'(x) = \dfrac{ {x}^{2} }{(x + 1) {(x + 2)}^{2} }[/tex]
☼︎~Now, as
[tex]\longmapsto\rm {\: {x}^{2} \geqslant 0x2⩾0}[/tex]
[tex]\longmapsto\sf\:{(x+2)}^{2} > 0[/tex]
[tex]\longmapsto\sf\: x + 1 > 0x+1>0[/tex]
☼︎~So,
[tex]\sf\longmapsto \:\rm \: f'(x) = \dfrac{ {x}^{2} }{(x + 1) {(x + 2)}^{2} } \geqslant 0[/tex]
☼︎~Therefore,
- f'(x) is always increasing when x > - 1
[tex]{ \underline { \underline{ \bold {\rule{200cm}{0.4cm}}}}}[/tex]
Additional Information:-
[tex]\boxed{\begin{array}{c|c} \bf f(x) & \tt \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 {e}^{x} & \sf {e}^{x} \end{array}}[/tex]
[tex]{ \underline { \underline{ \bold {\rule{200cm}{0.3cm}}}}}[/tex]
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