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y=(\sin x+\cos x)^x \text {, then } \frac{d y}{d x} \text { is }

Option: 1

y(sinx + cosx)


Option: 2

\left\{\frac{\log y}{x}+\frac{x(\cos x-\sin x)}{\sin x+\cos x}\right\}


Option: 3

y\left(\frac{\log y}{x}+y^2\right)


Option: 4

none of these 

 


Answers (1)

best_answer

y = (sin x + cos x)x

log y = x log (sin x + cos x) 

\begin{aligned} & \frac{1}{y} \cdot \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] \end{aligned}

Hence (B) is correct answer.

Posted by

Gaurav

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