Math Problem Statement

Use chain rule

Use chain rule to differentiate the following functions: 13. y = (2(x + 4)) / ((x + 4)^2 - 1), 14. y = 1 / (1 + sqrt(1 - x))^2

Solution

Let's solve both problems using the chain rule.

Problem 13:

Given: y=2(x+4)(x+4)21y = \frac{2(x + 4)}{(x + 4)^2 - 1} To differentiate this, we'll apply the quotient rule along with the chain rule.

Step-by-step differentiation:

Let: u=2(x+4),v=(x+4)21u = 2(x + 4), \quad v = (x + 4)^2 - 1 The quotient rule states: ddx(uv)=vdudxudvdxv2\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

Now we calculate dudx\frac{du}{dx} and dvdx\frac{dv}{dx}.

  1. dudx=ddx(2(x+4))=2\frac{du}{dx} = \frac{d}{dx} \left( 2(x + 4) \right) = 2
  2. dvdx=ddx((x+4)21)=2(x+4)\frac{dv}{dx} = \frac{d}{dx} \left( (x + 4)^2 - 1 \right) = 2(x + 4)

Now, applying the quotient rule: dydx=((x+4)21)(2)(2(x+4))(2(x+4))((x+4)21)2\frac{dy}{dx} = \frac{\left( (x + 4)^2 - 1 \right)(2) - \left( 2(x + 4) \right)(2(x + 4))}{\left( (x + 4)^2 - 1 \right)^2} Simplify: dydx=2((x+4)21)4(x+4)2((x+4)21)2\frac{dy}{dx} = \frac{2 \left( (x + 4)^2 - 1 \right) - 4(x + 4)^2}{\left( (x + 4)^2 - 1 \right)^2} This can be further simplified, but we'll leave it here for now.


Problem 14:

Given: y=1(1+1x)2y = \frac{1}{(1 + \sqrt{1 - x})^2} This requires the chain rule because of the square and the square root.

Step-by-step differentiation:

Rewrite the function as: y=(1+1x)2y = (1 + \sqrt{1 - x})^{-2}

Now, apply the chain rule.

  1. First, let u=1+1xu = 1 + \sqrt{1 - x}. Then y=u2y = u^{-2}.

  2. Differentiate y=u2y = u^{-2} with respect to uu: dydu=2u3\frac{dy}{du} = -2u^{-3}

  3. Now differentiate u=1+1xu = 1 + \sqrt{1 - x} with respect to xx: dudx=ddx(1+(1x)1/2)=12(1x)1/2(1)=121x\frac{du}{dx} = \frac{d}{dx} \left( 1 + (1 - x)^{1/2} \right) = \frac{1}{2}(1 - x)^{-1/2} \cdot (-1) = \frac{-1}{2\sqrt{1 - x}}

Now, combine the two results using the chain rule: dydx=dydududx=2u3121x\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -2u^{-3} \cdot \frac{-1}{2\sqrt{1 - x}}

Substitute u=1+1xu = 1 + \sqrt{1 - x} back into the expression: dydx=2(1+1x)321x\frac{dy}{dx} = \frac{2}{(1 + \sqrt{1 - x})^3 \cdot 2\sqrt{1 - x}} This simplifies to: dydx=1(1+1x)31x\frac{dy}{dx} = \frac{1}{(1 + \sqrt{1 - x})^3 \cdot \sqrt{1 - x}}


Would you like a more detailed step in any part?

Here are some related questions you might want to explore:

  1. How do you simplify expressions after applying the quotient rule?
  2. What is the best strategy for differentiating complicated functions involving fractions?
  3. Can the chain rule be combined with implicit differentiation?
  4. How can we handle higher-order derivatives of these functions?
  5. What role do critical points play in determining the behavior of these functions?

Tip: When using the chain rule, carefully identify each "inner" and "outer" function to differentiate correctly.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Quotient Rule
Radicals

Formulas

Quotient Rule: (u/v)' = (v * du/dx - u * dv/dx) / v^2
Chain Rule: dy/dx = dy/du * du/dx

Theorems

Quotient Rule
Chain Rule

Suitable Grade Level

Grades 11-12