Advertisements
Advertisements
Question
If \[y = \left( \sin x - \cos x \right)^{\sin x - \cos x} , \frac{\pi}{4} < x < \frac{3\pi}{4}, \text{ find} \frac{dy}{dx}\] ?
Advertisements
Solution
\[\text{ We have, y } = \left( \sin x - \cos x \right)^\left( \sin x - \cos x \right) \] ...(i)
Taking log on both sides,
\[\log y = \log \left( \sin x - \cos x \right)^\left( \sin x - \cos x \right) \]
\[ \Rightarrow \log y = \left( \sin x - \cos x \right) \log\left( \sin x - \cos x \right)\]
\[\Rightarrow \frac{1}{y}\frac{dy}{dx} = \log\left( \sin x - \cos x \right)\frac{d}{dx}\left( \sin x - \cos x \right) + \left( \sin x - \cos x \right)\frac{d}{dx}\log\left( \sin x - \cos x \right) \left[\text{ using product rule } \right]\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \log\left( \sin x - \cos x \right)\left( \cos x + \sin x \right) + \frac{\left( \sin x - \cos x \right)}{\left( \sin x - \cos x \right)}\frac{d}{dx}\left( \sin x - \cos x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \left( \cos x + \sin x \right) \log\left( \sin x - \cos x \right) + \left( \cos x + \sin x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \left( \cos x + \sin x \right)\left[ 1 + \log\left( \sin x - \cos x \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ \left( \cos x + \sin x \right)\left\{ 1 + \log\left( \sin x - \cos x \right) \right\} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \left( \sin x - \cos x \right)^\left( \sin x - \cos x \right) \left[ \left( \cos x + \sin x \right)\left\{ 1 + \log\left( \sin x - \cos x \right) \right\} \right] \left[ \text{ using equation } \left( i \right) \right]\]
