Math Problem Statement

Use the chain rule formula to find the derivative of the following functions. Express your final answers in terms of the variable x

Use the chain rule formula to find the derivative of the following functions: 3. y = (u - 4)^(5/2) where u = x^4 + 4, 4. y = 3u - 2 where u = 3x, 5. y = u^2 + 3u - 7 where u = 2x + 1. Express your final answers in terms of the variable x.

Solution

Let's go through each problem step by step, using the chain rule to differentiate the given functions. We will express all the final answers in terms of xx.

Problem 3:

y=(u4)52whereu=x4+4y = (u - 4)^{\frac{5}{2}} \quad \text{where} \quad u = x^4 + 4

To differentiate yy with respect to xx, we need to apply the chain rule.

First, differentiate yy with respect to uu: dydu=52(u4)32\frac{dy}{du} = \frac{5}{2} (u - 4)^{\frac{3}{2}}

Next, differentiate uu with respect to xx: dudx=4x3\frac{du}{dx} = 4x^3

Now, apply the chain rule: dydx=dydududx=52(u4)324x3\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{5}{2} (u - 4)^{\frac{3}{2}} \cdot 4x^3

Substitute u=x4+4u = x^4 + 4 into the expression: dydx=52((x4+4)4)324x3\frac{dy}{dx} = \frac{5}{2} \left( (x^4 + 4) - 4 \right)^{\frac{3}{2}} \cdot 4x^3 dydx=10x3(x4)32=10x3x6=10x9\frac{dy}{dx} = 10x^3 (x^4)^{\frac{3}{2}} = 10x^3 \cdot x^6 = 10x^9

Problem 4:

y=3u2whereu=3xy = 3u - 2 \quad \text{where} \quad u = 3x

First, differentiate yy with respect to uu: dydu=3\frac{dy}{du} = 3

Next, differentiate uu with respect to xx: dudx=3\frac{du}{dx} = 3

Now, apply the chain rule: dydx=dydududx=33=9\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3 \cdot 3 = 9

Problem 5:

y=u2+3u7whereu=2x+1y = u^2 + 3u - 7 \quad \text{where} \quad u = 2x + 1

First, differentiate yy with respect to uu: dydu=2u+3\frac{dy}{du} = 2u + 3

Next, differentiate uu with respect to xx: dudx=2\frac{du}{dx} = 2

Now, apply the chain rule: dydx=dydududx=(2u+3)2\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (2u + 3) \cdot 2 Substitute u=2x+1u = 2x + 1: dydx=(2(2x+1)+3)2=(4x+2+3)2=(4x+5)2\frac{dy}{dx} = (2(2x + 1) + 3) \cdot 2 = (4x + 2 + 3) \cdot 2 = (4x + 5) \cdot 2 dydx=8x+10\frac{dy}{dx} = 8x + 10


Do you need further clarifications on any of the steps?

Here are 5 related questions:

  1. How does the chain rule apply to multi-variable functions?
  2. What are some real-life applications of the chain rule?
  3. Can you explain why we differentiate yy with respect to uu first in these problems?
  4. What would happen if uu was a more complex function of xx?
  5. How does the chain rule change when applied to implicit differentiation?

Tip: Always start by identifying the outer and inner functions when using the chain rule!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Algebra

Formulas

Chain rule: dy/dx = dy/du * du/dx

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12 (Advanced Calculus)