Question

Find dy/dx of y=(logsinx/logcosx)(logcosx/logsinx)^-1

Answer 1

$\underline{\textbf{\large{Step-by-step explanation:}}}$

$y \: = \: \frac{(log \: sinx)}{(log \: cosx)} \times {( \frac{(log \: cosx)}{(log \: sinx)}) }^{ - 1}$

$= \frac{(log \: sinx) \: }{(log \: cosx)} \times \frac{1}{ \frac{(log \: cosx)}{(log \: sinx)} }$

$= ( \frac{(log \: sinx)}{(log \: cosx)} ) \times ( \frac{(log \: sinx)}{(log \: cosx)} )$

$= \frac{ {(log \: sinx)}^{2} }{ {(log \: cosx)}^{2} }$

$= {( \frac{log \: sinx}{log \: cosx}) }^{2}$

$= {\: ( log \: sinx \: - log \: cosx \:) }^{2}$

$\underline{\textbf{Differentiation wrt x : }}$

$\frac{dy}{dx} = \frac{d}{dx} {(log sinx - log cosx) }^{2} </h3><h3>$

$= 2(log \: sinx \: - log \: cosx)\\ \times [ ( \frac{1}{sin x} )(cos x) - ( \frac{1}{cosx} )( - sinx)]$

$= 2(log \: sinx \: - log \: cosx)(cotx \: + tanx)$

$\boxed{\textbf{dy / dx = </p><p> 2(log sinx - log cosx) (cotx + tanx)}}$

Answer 2

Step-by-step explanation:

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