Question
Find $\frac{d y}{d x}$ if : $x=\cos (\log t), y=\log (\cos t)$

Answer

Given, $y=\log (\cos t)$
Differentiate $w . r . t . t$
$
\frac{d y}{d t}=\frac{d}{d t}[\log (\cos t)]=\frac{1}{\cot t} \cdot \frac{d}{d t}(\cos t)=\frac{1}{\cot t}(-\sin t) \quad \therefore \frac{d y}{d t}=-\tan t
$
And, $x=\cos (\log t)$
Differentiate w.r.t.t
$
\frac{d x}{d t}=\frac{d}{d t}[\cos (\log t)]=-\sin (\log t) \cdot \frac{d}{d t}(\log t)=-\frac{\sin (\log t)}{t} \quad \therefore \frac{d x}{d t}=-\frac{\sin (\log t)}{t} .
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-\tan t}{-\frac{\sin (\log t)}{t}} \ldots[$ From (I) and (II) ]
$
\therefore \quad \frac{d y}{d x}=\frac{t \cdot \tan t^t}{\sin (\log t)}
$

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