d/dx[x^{ x} + ( x)^x] =

Correct Answer is : xsin⁡xx^{ x}xsinx [sin⁡xx+cos⁡xlog⁡x]≤ft[ x/x + x x ][xsinx+cosxlogx] + (sin⁡x)x( x)^x(sinx)x [x tan x + log (sin x)]

x^{ x} ≤ft[ x/x + x x ] + ( x)^x [x tan x + log (sin x)]

x^{ x} [x tan x + log (sin x)] + ( x)^x x^{ x} ≤ft[ x/x + ( x) ]

x^{ x} ≤ft[ x/ x + x x ] + ( x)^x [x cot x + log (sin x)]

x^{ x} ≤ft[ x/x + x x ] + ( x)^x [x cot x + log (cos x)]

Correct Answer is : xsin⁡xx^{ x}xsinx [sin⁡xx+cos⁡xlog⁡x]≤ft[ x/x + x x ][xsinx+cosxlogx] + (sin⁡x)x( x)^x(sinx)x [x tan x + log (sin x)]

Test Index

AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Section: Mathematics

Question : 66 of 160

Marks: +1, -0

\frac{d}{dx}[x^{\sin x} + (\sin x)^x]

$x^{\sin x}$ $\left[ \frac{\sin x}{x} + \cos x \log x \right]$ + $(\sin x)^x$ [x tan x + log (sin x)]
$x^{\sin x}$ [x tan x + log (sin x)] + $(\sin x)^x$ $x^{\sin x}$ $\left[ \frac{\sin x}{x} + \log(\sin x) \right]$
$x^{\sin x}$ $\left[ \frac{x}{\sin x} + \cos x \log x \right]$ + $(\sin x)^x$ [x cot x + log (sin x)]
$x^{\sin x}$ $\left[ \frac{\sin x}{x} + \sin x \log x \right]$ + $(\sin x)^x$ [x cot x + log (cos x)]

Solution:

Let

U=x^{\sin x}

and

V=(\sin x)^x

\log U=\sin x \log x

\frac{1}{U}\frac{dU}{dx} = \frac{\sin x}{x} + \cos x \log x

\frac{dU}{dx} = x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right]

Similarly,

\log V = x \log \sin x

\frac{1}{V}\frac{dV}{dx} = \log \sin x + x \cot x

\frac{dV}{dx} = (\sin x)^x \left[ \log \sin x + x \cot x \right]

Therefore,

\frac{d}{dx} \left[ x^{\sin x} + (\sin x)^x \right] = \frac{dU}{dx} + \frac{dV}{dx}

= x^{\sin x} \left[ \frac{\sin x}{x} + \cos x \log x \right]

+ (\sin x)^x \left[ \log \sin x + x \cot x \right]

Thus, no option is correct.

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