CUET Preparation Today
$y=(\sin x+\cos x)^x$, then $\frac{d y}{d x}$ is |
$y(\sin x+\cos x)$ y $\left\{\frac{\log y}{x}+\frac{x(\cos x-\sin x)}{\sin x+\cos x}\right\}$ $y\left(\frac{\log y}{x}+y^2\right)$ none of these |
y $\left\{\frac{\log y}{x}+\frac{x(\cos x-\sin x)}{\sin x+\cos x}\right\}$ |
$y=(\sin x+\cos x)^{x}$ $\log y=x \log (\sin x+\cos x)$ $\frac{1}{y} . \frac{d y}{d x}=\log (\sin x+\cos x)+\frac{x \times(\cos x-\sin x)}{\sin x+\cos x}$ $\frac{d y}{d x}=(\sin x+\cos x)^{x}\left[\frac{x(\cos x-\sin x}{\sin x+\cos x}+\log (\sin x+\cos x)\right] = y \left\{\frac{\log y}{x}+\frac{x(\cos x-\sin x)}{\sin x+\cos x}\right\}$ Hence (2) is correct answer. |
