Answer :

Let u = x^{sin} ^{x}

Taking log both sides,

⇒ log u = sin x log x

On differentiating,

Let, v = sin (x^{x})

Let, t = x^{x}

⇒ log t = x log x

On differentiating,

Using (2),

As, y = u + v

So,

From (1) and (3),

**OR**

y = log (1 + 2t^{2} + t^{4})

Applying the formula (a + b)^{2} = a^{2} + b^{2} + 2ab in (1 + 2t^{2} + t^{4}).

⇒ y = log (1+t^{2})^{2}

As log x^{n} = n log x

⇒ y = 2 log (1+t^{2})

On differentiating,

x = tan^{-1} t

On differentiating,

Now,

Again differentiating,

=4(1+t^{2})

Therefore,

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