Math Problem Statement

Given two functions: a) y = 7x^4 * sqrt(x^3 + 5) and b) y = (2x + 7)^5 / x^2, find the derivatives.

Solution

Let's break down the two expressions in the image and find solutions:

a) y=7x4x3+5y = 7x^4 \cdot \sqrt{x^3 + 5}

Here, we are given the function y=7x4x3+5y = 7x^4 \cdot \sqrt{x^3 + 5}, and I assume the question is asking for the derivative dydx\frac{dy}{dx}.

We will use the product rule combined with the chain rule for differentiation.

Steps:

  1. The first term is 7x47x^4, and its derivative is 28x328x^3.
  2. The second term is x3+5\sqrt{x^3 + 5}, which is equal to (x3+5)1/2(x^3 + 5)^{1/2}.
    • Applying the chain rule to differentiate (x3+5)1/2(x^3 + 5)^{1/2}, we get: ddx(x3+5)1/2=12(x3+5)1/23x2=3x22x3+5.\frac{d}{dx} (x^3 + 5)^{1/2} = \frac{1}{2}(x^3 + 5)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 5}}.

Using the product rule: dydx=ddx(7x4)(x3+5)+(7x4)ddx(x3+5)\frac{dy}{dx} = \frac{d}{dx} \left( 7x^4 \right) \cdot \left( \sqrt{x^3 + 5} \right) + \left( 7x^4 \right) \cdot \frac{d}{dx} \left( \sqrt{x^3 + 5} \right)

So, dydx=28x3x3+5+7x43x22x3+5.\frac{dy}{dx} = 28x^3 \cdot \sqrt{x^3 + 5} + 7x^4 \cdot \frac{3x^2}{2\sqrt{x^3 + 5}}. This is the derivative for part (a).

b) y=(2x+7)5x2y = \frac{(2x + 7)^5}{x^2}

Here, we are given the function y=(2x+7)5x2y = \frac{(2x + 7)^5}{x^2}, and again, I assume the question is asking for the derivative dydx\frac{dy}{dx}.

We will use the quotient rule and the chain rule to differentiate this expression.

Steps:

Define: u=(2x+7)5,v=x2.u = (2x + 7)^5, \quad v = x^2.

The quotient rule states: dydx=vdudxudvdxv2.\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}.

  1. u=(2x+7)5u = (2x + 7)^5, and using the chain rule: dudx=5(2x+7)42=10(2x+7)4.\frac{du}{dx} = 5(2x + 7)^4 \cdot 2 = 10(2x + 7)^4.
  2. v=x2v = x^2, and the derivative is: dvdx=2x.\frac{dv}{dx} = 2x.

Now, applying the quotient rule: dydx=x210(2x+7)4(2x+7)52xx4.\frac{dy}{dx} = \frac{x^2 \cdot 10(2x + 7)^4 - (2x + 7)^5 \cdot 2x}{x^4}.

Simplifying: dydx=10x2(2x+7)42x(2x+7)5x4.\frac{dy}{dx} = \frac{10x^2(2x + 7)^4 - 2x(2x + 7)^5}{x^4}. We can factor out 2x(2x+7)42x(2x + 7)^4: dydx=2x(2x+7)4[5x(2x+7)]x4.\frac{dy}{dx} = \frac{2x(2x + 7)^4 \left[ 5x - (2x + 7) \right]}{x^4}. Finally, simplify the expression in the brackets: 5x(2x+7)=3x7,5x - (2x + 7) = 3x - 7, which gives the final expression: dydx=2x(2x+7)4(3x7)x4.\frac{dy}{dx} = \frac{2x(2x + 7)^4 (3x - 7)}{x^4}. Canceling out xx gives: dydx=2(2x+7)4(3x7)x3.\frac{dy}{dx} = \frac{2(2x + 7)^4 (3x - 7)}{x^3}.


Would you like further clarification on any of these steps? Feel free to ask questions.

Here are 5 related questions to explore:

  1. What is the general product rule for differentiation?
  2. How can the chain rule be applied to more complex nested functions?
  3. How do you simplify expressions after applying the quotient rule?
  4. Can the second derivative d2ydx2\frac{d^2y}{dx^2} be computed for these functions?
  5. How do you approach integration for functions like the ones given in parts (a) and (b)?

Tip: Always simplify your final derivative expressions to ensure they are in their cleanest form.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule
Quotient Rule
Chain Rule

Formulas

Product Rule: (f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
Quotient Rule: (f(x)/g(x))' = (g(x)f'(x) - f(x)g'(x)) / (g(x))^2
Chain Rule: (f(g(x)))' = f'(g(x)) * g'(x)

Theorems

Product Rule
Quotient Rule
Chain Rule

Suitable Grade Level

Undergraduate (Calculus I)