Math Problem Statement

In Exercises 61–72, find the derivative of the function.

Solution

Let's find the derivatives for each of these functions using the appropriate differentiation rules, such as the chain rule and quotient rule.


Problem 61

Given: y=(7x+3)4y = (7x + 3)^4

To differentiate, use the chain rule:

  1. Let u=7x+3u = 7x + 3, then y=u4y = u^4.
  2. The derivative of yy with respect to uu is dydu=4u3\frac{dy}{du} = 4u^3.
  3. The derivative of uu with respect to xx is dudx=7\frac{du}{dx} = 7.

Using the chain rule: dydx=dydududx=4(7x+3)37=28(7x+3)3\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4(7x + 3)^3 \cdot 7 = 28(7x + 3)^3


Problem 62

Given: y=(x26)3y = (x^2 - 6)^3

Use the chain rule here as well:

  1. Let u=x26u = x^2 - 6, then y=u3y = u^3.
  2. The derivative of yy with respect to uu is dydu=3u2\frac{dy}{du} = 3u^2.
  3. The derivative of uu with respect to xx is dudx=2x\frac{du}{dx} = 2x.

Using the chain rule: dydx=dydududx=3(x26)22x=6x(x26)2\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3(x^2 - 6)^2 \cdot 2x = 6x(x^2 - 6)^2


Problem 63

Given: y=1(x2+5)3y = \frac{1}{(x^2 + 5)^3}

Rewrite the function as: y=(x2+5)3y = (x^2 + 5)^{-3}

Differentiate using the chain rule:

  1. Let u=x2+5u = x^2 + 5, then y=u3y = u^{-3}.
  2. The derivative of yy with respect to uu is dydu=3u4\frac{dy}{du} = -3u^{-4}.
  3. The derivative of uu with respect to xx is dudx=2x\frac{du}{dx} = 2x.

Using the chain rule: dydx=dydududx=3(x2+5)42x=6x(x2+5)4\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -3(x^2 + 5)^{-4} \cdot 2x = -\frac{6x}{(x^2 + 5)^4}


Problem 64

Given: f(x)=1(5x+1)2f(x) = \frac{1}{(5x + 1)^2}

Rewrite the function as: f(x)=(5x+1)2f(x) = (5x + 1)^{-2}

Differentiate using the chain rule:

  1. Let u=5x+1u = 5x + 1, then f(x)=u2f(x) = u^{-2}.
  2. The derivative of ff with respect to uu is dfdu=2u3\frac{df}{du} = -2u^{-3}.
  3. The derivative of uu with respect to xx is dudx=5\frac{du}{dx} = 5.

Using the chain rule: dfdx=dfdududx=2(5x+1)35=10(5x+1)3\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = -2(5x + 1)^{-3} \cdot 5 = -\frac{10}{(5x + 1)^3}


Would you like further details or explanations on any specific step?

Related Questions:

  1. How can the product rule be combined with the chain rule in differentiation?
  2. What are the applications of derivatives in real-world problems?
  3. How does implicit differentiation differ from the chain rule?
  4. Can you differentiate composite functions that involve trigonometric terms?
  5. How do we handle higher-order derivatives for similar types of functions?

Tip:

Always simplify the expression of the derivative as much as possible to make it easier for further calculations or applications.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Chain Rule
Quotient Rule

Formulas

Chain rule: dy/dx = (dy/du) * (du/dx)
Quotient rule: (f/g)' = (f'g - fg') / g^2

Theorems

Chain Rule
Quotient Rule

Suitable Grade Level

Grade 12 or College Calculus