For constant a, d/dx(x^{x}+x^{a}+a^{x}+a^{a}) is

Correct Answer is : xx(1+log⁡x)+axa−1+axlog⁡ax^{x}(1+ x)+a x^{a-1}+a^{x} axx(1+logx)+axa−1+axloga

x^{x}(1+ x)+a^{a}(1+ a)+a x^{a-1}

Correct Answer is : xx(1+log⁡x)+axa−1+axlog⁡ax^{x}(1+ x)+a x^{a-1}+a^{x} axx(1+logx)+axa−1+axloga

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KCET 2021 Math Solved Paper

Question : 39 of 60

Marks: +1, -0

For constant

a, \frac{d}{dx}(x^{x}+x^{a}+a^{x}+a^{a})

Solution:

\frac{d}{dx}(x^{x}+x^{a}+a^{x}+a^{a})

=\frac{d}{dx}(x^{x})+\frac{d}{dx}(x^{a})+\frac{d}{dx}(a^{x})+\frac{d}{dx}(a^{a})

=\frac{d}{dx}(y)+a x^{a-1}+a^{x}\log a+0

...(i)

y=x^{x}

[\text{let} x^{x}=y]

Taking log on both sides,

\log y=x \log x

On differentiating w.r.t.

x

, we get

\frac{1}{y} \frac{dy}{dx}=x \times \frac{1}{x}+\log x

⇒

\frac{dy}{dx}=x^{x}(1+\log x)

Substitute the value of

\frac{dy}{dx}

in Eq. (i),

\frac{d}{dx}(x^{x}+x^{a}+a^{x}+a^{a})

=x^{x}(1+\log x)+a x^{a-1}+a^{x}\log a

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