TIFR CSE 2018 | Part A | Question: 5

Question

Which of the following is the derivative of $f(x)=x^{x}$ when $x>0$ ?

• A. $x^{x}$
• B. $x^{x} \ln \;x$
• C. $x^{x}+x^{x}\ln\;x$
• D. $(x^{x}) (x^{x}\ln\;x)$
• E. $\text{None of the above; function is not differentiable for }x>0$

Comments

Option C

Answer 1

Let $y=x^{x}$

Taking natural log on both the sides, we get

$\ln\ y=x\ln\ x$

Differentiate both the sides wrt to x we get

$\frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}$ = $(1) \ln \ x +x\frac{1}{x} $ (Product Rule)

$\frac{\mathrm{d}y}{\mathrm{d}x}$ = $y \ln\ x + y$

Substitute $y = x^{x}$

$\frac{\mathrm{d}y }{\mathrm{d} x}$=$\ x^{x}\ln\ x + x^{x}$

Hence, option C is correct answer.

Comments

when we are differentiating with x then logy becomes constant with respect to x then how can you differentiate logy?

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@Salla shivateja log(y) is a function of x, as we assumed y=x^x