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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability2 Marks
Question
Differentiate the $\frac{{\cos x}}{{\log x}},x > 0$ w.r.t. x.
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Answer

Let $y = \frac{{\cos x}}{{\log x}}$ 

$\therefore \frac{{dy}}{{dx}} = \frac{{\log x\frac{d}{{dx}}\left( {\cos x} \right) - \cos x\frac{d}{{dx}}\left( {\log x} \right)}}{{{{\left( {\log x} \right)}^2}}}$ [By quotient rule]

$= \frac{{\log x\left( { - \sin x} \right) - \cos x.\frac{1}{x}}}{{{{\left( {\log x} \right)}^2}}}$

$= \frac{{ - \left( {\sin x\log x + \frac{{\cos x}}{x}} \right)}}{{{{\left( {\log x} \right)}^2}}}$

$= \frac{{ - \left( {x\sin x\log x + \cos x} \right)}}{{x{{\left( {\log x} \right)}^2}}}$

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