Question

find d/dx=
$log_{e(x + 1 \div \sqrt{x - 2)} }(?)$


​

Answer 1

We know,

• $\displaystyle\sf {\log\left (\dfrac {a}{b}\right)=\log a-\log b}$
• $\displaystyle\sf {\log (a^b)=b\log a}$

Then we get,

$\displaystyle\sf{\longrightarrow\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {d}{dx}\left [\log(x+1)-\dfrac {1}{2}\log (x-2)\right]}$

Since $\displaystyle\sf {\dfrac {d}{dx}\left [\log (f(x))\right]=\dfrac {f'(x)}{f(x)},}$

$\displaystyle\sf{\longrightarrow\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {1}{x+1}-\dfrac {1}{2}\cdot\dfrac {1}{x-2}}$

$\displaystyle\sf{\longrightarrow\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {1}{x+1}-\dfrac {1}{2(x-2)}}$

$\displaystyle\sf{\longrightarrow\underline {\underline {\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {x-5}{2(x+1)(x-2)}}}}$

Answer 2

We know,

\displaystyle\sf {\log\left (\dfrac {a}{b}\right)=\log a-\log b}log(

b

a

)=loga−logb

\displaystyle\sf {\log (a^b)=b\log a}log(a

b

)=bloga

Then we get,

\displaystyle\sf{\longrightarrow\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {d}{dx}\left [\log(x+1)-\dfrac {1}{2}\log (x-2)\right]}⟶

dx

d

[log(

x−2

x+1

)]=

dx

d

[log(x+1)−

2

1

log(x−2)]

Since \displaystyle\sf {\dfrac {d}{dx}\left [\log (f(x))\right]=\dfrac {f'(x)}{f(x)},}

dx

d

[log(f(x))]=

f(x)

f

′

(x)

,

\displaystyle\sf{\longrightarrow\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {1}{x+1}-\dfrac {1}{2}\cdot\dfrac {1}{x-2}}⟶

dx

d

[log(

x−2

x+1

)]=

x+1

1

−

2

1

⋅

x−2

1

\displaystyle\sf{\longrightarrow\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {1}{x+1}-\dfrac {1}{2(x-2)}}⟶

dx

d

[log(

x−2

x+1

)]=

x+1

1

−

2(x−2)

1

\displaystyle\sf{\longrightarrow\underline {\underline {\dfrac {d}{dx}\left [\log\left (\dfrac {x+1}{\sqrt {x-2}}\right)\right]=\dfrac {x-5}{2(x+1)(x-2)}}}}⟶

dx

d

[log(

x−2

x+1

)]=

2(x+1)(x−2)

x−5