Solve this


If $y=(\sin x-\cos x)^{\sin x-\cos x}, \frac{\pi}{4}


Here, $y=(\sin x-\cos x)^{(\sin x-\cos x)}$                     .......(1)

Taking log on both sides,

$\log y=\log (\sin x-\cos x)^{(\sin x-\cos x)}$

$\log y=(\sin x-\cos x) \log (\sin x-\cos x)$

Differentiating it with respect to $x$ using product rule, chain rule,

$\frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \frac{d}{d x}(\sin x-\cos x)+(\sin x-\cos x) \frac{d}{d x} \log (\sin x-\cos x)$

$\frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \times(\cos x+\sin x)+\frac{(\sin x-\cos x)}{(\sin x-\cos x)} \frac{d}{d x}(\sin x-\cos x)$

$\frac{1}{y} \frac{d y}{d x}=(\cos x+\sin x) \log (\sin x-\cos x)+(\cos x+\sin x)$

$\frac{1}{y} \frac{d y}{d x}=(\cos x+\sin x)(1+\log (\sin x-\cos x))$

$\frac{d y}{d x}=y[(\cos x+\sin x)(1+\log (\sin x-\cos x))]$

Using (i),

$\frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}[(\cos x+\sin x)(1+\log (\sin x-\cos x))]$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now