Math Problem Statement

Question content area top Part 1 Find StartFraction d squared y Over dx squared EndFraction ​, where StartRoot y EndRootplus2xyequals1.

just give me the exact formula for what the second derivative equals

Solution

Given the equation y+2xy=1\sqrt{y} + 2xy = 1, we need to find d2ydx2\frac{d^2 y}{dx^2}.

First, let's rewrite the equation in a more convenient form: y+2xy=1\sqrt{y} + 2xy = 1

We differentiate both sides with respect to xx to find dydx\frac{dy}{dx}: ddx(y)+ddx(2xy)=ddx(1)\frac{d}{dx}(\sqrt{y}) + \frac{d}{dx}(2xy) = \frac{d}{dx}(1)

Using the chain rule and product rule: 12ydydx+2y+2xdydx=0\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} = 0

Let's isolate dydx\frac{dy}{dx}: 12ydydx+2xdydx=2y\frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} + 2x \frac{dy}{dx} = -2y (12y+2x)dydx=2y\left(\frac{1}{2\sqrt{y}} + 2x\right) \frac{dy}{dx} = -2y dydx=2y12y+2x\frac{dy}{dx} = \frac{-2y}{\frac{1}{2\sqrt{y}} + 2x}

Now, let's simplify this: dydx=2y1+4xy2y\frac{dy}{dx} = \frac{-2y}{\frac{1 + 4x\sqrt{y}}{2\sqrt{y}}} dydx=2y2y1+4xy\frac{dy}{dx} = \frac{-2y \cdot 2\sqrt{y}}{1 + 4x\sqrt{y}} dydx=4yy1+4xy\frac{dy}{dx} = \frac{-4y\sqrt{y}}{1 + 4x\sqrt{y}}

Next, we differentiate dydx\frac{dy}{dx} with respect to xx to find d2ydx2\frac{d^2y}{dx^2}: ddx(4yy1+4xy)\frac{d}{dx} \left( \frac{-4y\sqrt{y}}{1 + 4x\sqrt{y}} \right)

Using the quotient rule: d2ydx2=(1+4xy)ddx(4yy)(4yy)ddx(1+4xy)(1+4xy)2\frac{d^2 y}{dx^2} = \frac{(1 + 4x\sqrt{y}) \cdot \frac{d}{dx}(-4y\sqrt{y}) - (-4y\sqrt{y}) \cdot \frac{d}{dx}(1 + 4x\sqrt{y})}{(1 + 4x\sqrt{y})^2}

First, we find ddx(4yy)\frac{d}{dx}(-4y\sqrt{y}): ddx(4yy)=4(ddx(yy))\frac{d}{dx}(-4y\sqrt{y}) = -4 \left( \frac{d}{dx}(y\sqrt{y}) \right) Using the product rule and chain rule: ddx(yy)=ydydx+y12ydydx=ydydx+y2ydydx=(y+y2y)dydx=32ydydx\frac{d}{dx}(y\sqrt{y}) = \sqrt{y} \cdot \frac{dy}{dx} + y \cdot \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = \sqrt{y} \cdot \frac{dy}{dx} + \frac{y}{2\sqrt{y}} \cdot \frac{dy}{dx} = \left( \sqrt{y} + \frac{y}{2\sqrt{y}} \right) \frac{dy}{dx} = \frac{3}{2} \sqrt{y} \cdot \frac{dy}{dx}

So: ddx(4yy)=432ydydx=6ydydx\frac{d}{dx}(-4y\sqrt{y}) = -4 \cdot \frac{3}{2} \sqrt{y} \cdot \frac{dy}{dx} = -6 \sqrt{y} \cdot \frac{dy}{dx}

Next, we find ddx(1+4xy)\frac{d}{dx}(1 + 4x\sqrt{y}): ddx(1+4xy)=4(y+x12ydydx)=4y+2xydydx\frac{d}{dx}(1 + 4x\sqrt{y}) = 4 \left( \sqrt{y} + x \cdot \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \right) = 4\sqrt{y} + \frac{2x}{\sqrt{y}} \cdot \frac{dy}{dx}

Putting it all together: d2ydx2=(1+4xy)(6ydydx)(4yy)(4y+2xydydx)(1+4xy)2\frac{d^2 y}{dx^2} = \frac{(1 + 4x\sqrt{y})(-6 \sqrt{y} \cdot \frac{dy}{dx}) - (-4y\sqrt{y})(4\sqrt{y} + \frac{2x}{\sqrt{y}} \cdot \frac{dy}{dx})}{(1 + 4x\sqrt{y})^2}

Simplifying each term: =6ydydx24xydydx(16yy8xydydx)(1+4xy)2= \frac{-6 \sqrt{y} \cdot \frac{dy}{dx} - 24x y \cdot \frac{dy}{dx} - (-16 y \cdot y - 8xy \cdot \frac{dy}{dx})}{(1 + 4x\sqrt{y})^2} =6ydydx24xydydx+16y2+8xydydx(1+4xy)2= \frac{-6 \sqrt{y} \cdot \frac{dy}{dx} - 24xy \cdot \frac{dy}{dx} + 16y^2 + 8xy \cdot \frac{dy}{dx}}{(1 + 4x\sqrt{y})^2} =6ydydx16xydydx+16y2(1+4xy)2= \frac{-6 \sqrt{y} \cdot \frac{dy}{dx} - 16xy \cdot \frac{dy}{dx} + 16y^2}{(1 + 4x\sqrt{y})^2}

Finally, substitute dydx\frac{dy}{dx} back in to get the explicit formula for d2ydx2\frac{d^2 y}{dx^2}.

Would you like further details or have any questions?

  1. How do we apply the chain rule in differentiation?
  2. What is the quotient rule and how is it used in differentiation?
  3. How can implicit differentiation help solve complex equations?
  4. What are some common applications of second derivatives?
  5. How do product and chain rules differ in their applications?

Tip: Always simplify expressions step by step to avoid errors in differentiation.

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Chain Rule
Product Rule
Quotient Rule
Second Derivative

Formulas

\( \frac{d}{dx}(\sqrt{y}) + \frac{d}{dx}(2xy) = \frac{d}{dx}(1) \)
\( \frac{dy}{dx} = \frac{-4y\sqrt{y}}{1 + 4x\sqrt{y}} \)
Quotient Rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

Theorems

Chain Rule
Product Rule
Quotient Rule

Suitable Grade Level

Advanced High School or Early College Level (Grades 11-12 or College Calculus)