If y = (x^5/e^5), then d^2y/dx^2 is Option 1: (-5)/x^2 Option 2: (-20)/x^4 Option 3: x^3/e^2 Option 4: (-20x^3)/e^2

Answer: (-5)/x^2. Solution: y = 5 x - 5. dy/dx = 5/x. d^2y/dx^2 = -5/x^2.

Applied Mathematics

Calculus

Easy

applied

If $y = \log\left(\frac{x^5}{e^5}\right)$ , then $\frac{d^2y}{dx^2}$ is,

$\frac{-5}{x^2}$

$\frac{-20}{x^4}$

$\frac{x^3}{e^2}$

$\frac{-20x^3}{e^2}$

✅ Correct Option: 1

$y = 5\ln x - 5$ .

$\frac{dy}{dx} = \frac{5}{x}$ .

$\frac{d^2y}{dx^2} = -\frac{5}{x^2}$ .

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The Marginal Cost (MC), when $x = 10$ units is:

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The demand function for a commodity is $p = 45 + x - x^2$ , then the consumer's surplus at equilibrium price $p_0 = 25$ will be:

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The order and degree Of the differential equation $\frac{d^2 y}{dx^2} + 2 e^{-x} = 0$ , respectively are

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The value of the integral

$I = \int_{-1}^{1} (x + x^3 + x^5) dx$ is :

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A product costs the manufacturer ₹ 20 per unit. The demand function is given by $p(x) = 1000 - 20x$ , then the quantity for maximum profit is :

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Match List - I with List - II.

Choose the correct answer from the options given below :