Question

Solve .
$\frac{dy}{dx} \: of \: the \: given \: function.$
Please answer it.​

Answer 1

Let's simplify the expression step by step:

$\large \displaystyle \sf y = \sqrt{ \frac{ { \sin}^{ - 1}(logx) + {e}^{log( {x \frac{ - 1}{ \sqrt{2} }) }^{ \frac{}{ } } } }{log( {e}^{ \frac{1}{ \sqrt{x} } } ) - {sec}^{ - 1}( \sqrt{ {x}^{3} } ) } }$

$\large \displaystyle \sf y = \sqrt{ \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} }$

Now, let's use the chain rule to find dy/dx:

$\large \displaystyle \sf \frac{dy}{dx} = \frac{1}{2} \left( \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} \right)^{-\frac{1}{2}} \cdot \left( \frac{d}{dx} \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} \right)$

We can simplify the expression inside the derivative as follows:

$\large \displaystyle \sf \frac{d}{dx} \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} = \frac{ \frac{d}{dx} \left( { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} \right) \cdot (\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})) - (\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})) \cdot \frac{d}{dx} \left( { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} \right) }{(\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3}))^2}$

$\large \displaystyle \sf \frac{d}{dx} \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} = \frac{ \frac{1}{\sqrt{1-(logx)^2}} \cdot \frac{1}{x} + \frac{-1}{\sqrt{2}} \cdot (\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1-x^3}}) }{(\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3}))^2}$

Now we can substitute this expression and simplify further:

$\large \displaystyle \sf \frac{dy}{dx} = \frac{1}{2} \left( \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} \right)^{-\frac{1}{2}} \cdot \left( \frac{ \frac{1}{\sqrt{1-(logx)^2}} \cdot \frac{1}{x} + \frac{-1}{\sqrt{2}} \cdot (\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1-x^3}}) }{(\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3}))^2} \right)$

Now, we just need to simplify this expression:

$\large \displaystyle \sf \frac{dy}{dx} = \frac{1}{2} \left( \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} \right)^{-\frac{1}{2}} \cdot \frac{ \frac{1}{\sqrt{1-(logx)^2}} \cdot \frac{1}{x} + \frac{-1}{\sqrt{2}} \cdot (\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1-x^3}}) }{(\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3}))^2}$

Simplifying further, we get:

$\large \displaystyle \sf \frac{dy}{dx} = \frac{1}{2} \frac{ \frac{1}{\sqrt{1-(logx)^2}} \cdot \frac{1}{x} + \frac{-1}{\sqrt{2}} \cdot (\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1-x^3}}) }{ \sqrt{ \frac{ { \sin}^{ - 1}(logx) + x \frac{-1}{\sqrt{2}} }{\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3})} } \cdot (\frac{1}{\sqrt{x}} - \cos^{-1}(\sqrt{x^3}))^2 }$

This is the derivative of the given expression with respect to x.