Question

differentiate the (o) part w.r.t x

Required Answer :-

$\bigstar \boxed{\sf\dfrac{1}{x} \bigg[ \dfrac{1}{ log_{e}a }- \dfrac{ log_{e}a }{ (log_{e}x ) ^{2} } \bigg]} \bigstar$
Don't spam ~

​

Answer 1

Let ,

$\small \bf y = log_{x}(a) + log_{a}(x) + \bigg \{log_{x}(a) . log_{a}(x) \bigg \}$

___________________________

▪To Calculate :-

$\bf \pink{ \dfrac{dy}{dx} }$

___________________________

▪Formulae Used :-

$\maltese \: \: \: \bf log_{n}(m) = \dfrac{ log_{e}(m) }{ log_{e}(n) } \\ \\ \maltese \: \: \: \bf \frac{d}{dx} \: log \: x = \frac{1}{x} \\ \\ \maltese \: \: \: \bf \frac{d}{dx} \: \{ f(x) { \}}^{n} = n. \{f(x) \} {}^{n - 1} . \frac{d}{dx} \: f(x)$

___________________________

▪Solution :-

$\sf\text{ y }= log_{x}(a) + log_{a}(x) + \big\{log_{x}(a) . log_{a}(x) \big \} \\ \\ \sf= log_{x}(a) + log_{a}(x) + \bigg \{ \dfrac{ log_{e}(a) }{ log_{e}(x) } \times \frac{ log_{e}(x) }{ log_{e}(a) } \bigg \}$

$\implies \sf y = log_{x}(a) + log_{a}(x) + 1\\ \\ \implies \bf y = \frac{ log_{e}(a) }{ log_{e}(x) } + \frac{ log_{e}(x) }{ log_{e}(a) } + 1$

Let ,

$\sf A = \frac{ log_{e}(a) }{ log_{e}(e) } \\ \\ \sf B = \frac{ log_{e}(x) }{ log_{e}(a) }$

So ,

$\sf \dfrac{dy}{dx} = \dfrac{dA}{dx} + \dfrac{dB}{dx} + \dfrac{d1}{dx} \\ \\ \implies \sf \dfrac{dy}{dx} = \dfrac{dA}{dx} + \dfrac{dB}{dx} + 0 \\ \\ \implies\bf \dfrac{dy}{dx} = \dfrac{dA}{dx} + \dfrac{dB}{dx}\:\:\:\:.\:.\:.\:\{i\}$

We have ,

$\sf A = \frac{ log_{e}(a)}{ log_{e} (x)} \\ \\ \sf A = log_{e}(a) \: . \: \frac{1}{ log_{e}(x) }$

Differentiating both sides w.r.t x

$\sf \dfrac{dA }{dx} = log_{e}(a) \: .\: \{ log_{e}(x) { \}}^{ - 1} \\ \\ \sf = log_{e}(a) \:. \: ( - 1) \{{ log_{e}(x) \}}^{ - 2} . \frac{1}{x} \\ \\\large \purple{ \implies  \underline {\boxed{{\bf  \frac{dA}{dx}  =  - \frac{ log_{e}(a) } {x \{ { log_{e}(x) \}}^{2} }} }}}$

Also ,

$\sf B = \dfrac{ log_{e}(x) }{ log_{e}(a) } \\ \\ \sf B = \frac{1}{ log_{e}(a) } \: . \: log_{e}(x)$

Differentiating both sides w.r.t x

$\sf\dfrac{dB}{dx} = \dfrac{1}{ log_{e}(a) } \: .\: \dfrac{d}{dx} \: log_{e}(x) \\ \\ \large \purple{ \implies  \underline {\boxed{{\bf  \frac{dB}{dx}  =  \frac{1}{x( log_{e}(a) }} }}}$

Putting in equation {i} We Get ,

$\sf \dfrac{dy}{dx} = - \dfrac{ log_{e}(a) } {x \{ { log_{e}(x) \}}^{2} } \: + \: \dfrac{1}{x( log_{e}(a) } \\ \\ \purple{ \implies  \underline {\boxed{{\bf  \frac{dy}{dx}  = \dfrac{1}{x} \bigg[ \dfrac{1}{ log_{e}a }- \dfrac{ log_{e}a }{ (log_{e}x ) ^{2} } \bigg]} }}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

___________________________

Answer 2

Refer to the attachment for your answer .

Used Concepts :-

• Quotient rule of differentiation which states for any two functions u and v , d/dx ( u/v ) = v × du/dx - u × dv/dx/v²
• d/dx { ln ( x ) } = 1/x
• log_x y = ln ( y )/ln ( x )
• log_x y . log_y x = 1