Exams › JEE Main › Maths

If x = e^(y + e^(y +...)), x > 0, then dy/dx is

1. (1 + x)/x
2. 1/x
3. (1 - x)/x
4. x/(1 + x)

Correct answer: (1 - x)/x

Solution

The inner expression e^(y+...) equals x itself, so x=e^(y+x). Taking logs: ln x = y + x, hence y = ln x - x and dy/dx = 1/x - 1 = (1 - x)/x.

Related JEE Main Maths questions

• Evaluate the limit as x approaches 0: ((1+tan x)/(1+sin x))^(csc x).
• Evaluate the limit as x approaches 0 of sin([x−3]) divided by [x−3], where [.] denotes the greatest integer function.
• Evaluate the limit as x approaches 0: (sin(x⁴) − x⁴ cos(x⁴) + x²⁰) / (x⁴(e^(2x⁴) − 1 − 2x⁴)).
• Evaluate the limit lim_(n→∞)(n³-8)/(n³+8). What is its value?
• If lim_(x→∞) (1+(λ)/(x)+(μ)/(x²))^(2x)=e², which of the following values of λ and μ is correct?
• Consider the following statements: Statement 1: If lim_(x→α)(sin(f(x)))/(x-α), where f(x)=ax²+bx+c, exists as a finite nonzero number, then lim_(x→α)(1/f(x)-1)/(1/f(x)+1) does not exist. Statement 2: The limit lim_(x→α)(sin(f(x)))/(x-α) can be finite only in the indeterminate form 0/0.

⚔️ Practice JEE Main Maths free + battle 1v1 →