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Question:

$x^{x}+x^{a}+a^{x}+a^{a}$, for some fixed $a>0$ and $x>0$

Solution:

Let $y=x^{x}+x^{a}+a^{x}+a^{a}$

Also, let $x^{x}=u, x^{a}=v, a^{x}=w$, and $a^{a}=s$

$\therefore y=u+v+w+s$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x}$   ...(1)

$u=x^{x}$

$\Rightarrow \log u=\log x^{x}$

$\Rightarrow \log u=x \log x$

Differentiating both sides with respect to x, we obtain'

$\frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)$

$\Rightarrow \frac{d u}{d x}=u\left[\log x \cdot 1+x \cdot \frac{1}{x}\right]$

$\Rightarrow \frac{d u}{d x}=x^{x}[\log x+1]=x^{x}(1+\log x)$   ...(2)

$v=x^{a}$

$\therefore \frac{d v}{d x}=\frac{d}{d x}\left(x^{a}\right)$

$\Rightarrow \frac{d v}{d x}=a x^{a-1}$   ...(3)

$w=a^{x}$

$\Rightarrow \log w=\log a^{x}$

$\Rightarrow \log w=x \log a$

Differentiating both sides with respect to x, we obtain

$\frac{1}{w} \cdot \frac{d w}{d x}=\log a \cdot \frac{d}{d x}(x)$

$\Rightarrow \frac{d w}{d x}=w \log a$

$\Rightarrow \frac{d w}{d x}=a^{x} \log a$   ...(4)

$s=a^{a}$

Since $a$ is constant, $a^{a}$ is also a constant.

Since a is constant, aa is also a constant.

$\therefore \frac{d s}{d x}=0$     ....(5)

From (1), (2), (3), (4), and (5), we obtain

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

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

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