Question

If dy over dx equals x times the square root of the quantity 1 minus 4 times x end quantity comma what is d squared y over dx squared question mark (10 points)

d squared y over dx squared equals 1 minus 6 times x over the square root of the quantity 1 minus 4 times x end quantity

d squared y over dx squared equals 1 minus 2 times x over the square root of the quantity 1 minus 4 times x end quantity

d squared y over dx squared equals 1 minus 4 times x over the square root of the quantity 1 minus 4 times x end quantity

d squared y over dx squared equals 2 minus 4 times x over the square root of the quantity 1 minus 4 times x end quantity

Answer 1

Answer:

$\boxed{\dfrac{d^2y}{dx^2} = \dfrac{1-6x}{\sqrt{1-4x}}}$

Step-by-step explanation:

We are given the first derivative, dy/dx, of a function y in terms of x:

$\dfrac{dy}{dx} = x\sqrt{1-4x}$

and we are asked to solve for the second derivative:

$\dfrac{d^2y}{dx^2}$

To do this, we simply have to take the derivative of both sides of the given first derivative:

$\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right) = \dfrac{d}{dx}\!\left(x\sqrt{1-4x}\,\right)$

$\implies \dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(x\sqrt{1-4x}\,\right)$

We can differentiate the left side using the product rule, chain rule, and power rule:

• $(f\cdot g)' = f\cdot g' - g\cdot f'$
• $\left[ \dfrac{}{}f(g(x)) \dfrac{}{}\right]' = f'(g(x)) \cdot g'(x)$
• $(x^n)' = nx^{n-1}$

↓ applying the product rule

$\dfrac{d^2y}{dx^2} = x \cdot \dfrac{d}{dx}\!\left(\sqrt{1-4x}\,\right) + \sqrt{1-4x} \cdot \dfrac{d}{dx}(x)$

↓ applying the chain and power rules

$= x \left(\dfrac{1}{2}(1-4x)^{\left(\frac{1}2 - 1\right)} \cdot (-4) \right) + \sqrt{1-4x} \cdot 1$

$= \dfrac{-4x}{2\sqrt{1-4x}} + \sqrt{1-4x}$

$= \dfrac{-2x}{\sqrt{1-4x}} + \sqrt{1-4x}$

↓ combining both terms over a common denominator

$= \dfrac{-2x}{\sqrt{1-4x}} + \sqrt{1-4x}\left(\dfrac{\sqrt{1-4x}}{\sqrt{1-4x}}\right)$

$= \dfrac{-2x}{\sqrt{1-4x}} + \dfrac{1-4x}{\sqrt{1-4x}}$

$= \dfrac{-2x + (1-4x)}{\sqrt{1-4x}}$

$\boxed{\dfrac{d^2y}{dx^2} = \dfrac{1-6x}{\sqrt{1-4x}}}$