Math Problem Statement

Write the following function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x. y = ((x^2 / 4) + 3x - (3/x))^3

Solution

Let's begin by analyzing the expression provided in the image.

The function given is:

y=(x24+3x3x)3y = \left( \frac{x^2}{4} + 3x - \frac{3}{x} \right)^3

We are asked to write it in the form y=f(u)y = f(u) and u=g(x)u = g(x), and then find dydx\frac{dy}{dx} as a function of xx.

Step 1: Expressing the function in the form y=f(u)y = f(u) and u=g(x)u = g(x)

We define:

u=g(x)=x24+3x3xu = g(x) = \frac{x^2}{4} + 3x - \frac{3}{x}

Then, the function yy becomes:

y=f(u)=u3y = f(u) = u^3

Step 2: Differentiating y=f(u)y = f(u) and u=g(x)u = g(x)

We will use the chain rule to find dydx\frac{dy}{dx}:

dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Finding dydu\frac{dy}{du}:

Since y=u3y = u^3,

dydu=3u2\frac{dy}{du} = 3u^2

Finding dudx\frac{du}{dx}:

We need to differentiate u=x24+3x3xu = \frac{x^2}{4} + 3x - \frac{3}{x} with respect to xx:

dudx=ddx(x24)+ddx(3x)ddx(3x)\frac{du}{dx} = \frac{d}{dx} \left( \frac{x^2}{4} \right) + \frac{d}{dx} (3x) - \frac{d}{dx} \left( \frac{3}{x} \right)

  • ddx(x24)=2x4=x2\frac{d}{dx} \left( \frac{x^2}{4} \right) = \frac{2x}{4} = \frac{x}{2}
  • ddx(3x)=3\frac{d}{dx} (3x) = 3
  • ddx(3x)=3x2\frac{d}{dx} \left( \frac{3}{x} \right) = -\frac{3}{x^2}

Thus,

dudx=x2+3+3x2\frac{du}{dx} = \frac{x}{2} + 3 + \frac{3}{x^2}

Step 3: Substituting everything to find dydx\frac{dy}{dx}

Now, we substitute dydu=3u2\frac{dy}{du} = 3u^2 and dudx=x2+3+3x2\frac{du}{dx} = \frac{x}{2} + 3 + \frac{3}{x^2}:

dydx=3u2(x2+3+3x2)\frac{dy}{dx} = 3u^2 \cdot \left( \frac{x}{2} + 3 + \frac{3}{x^2} \right)

Finally, substitute u=x24+3x3xu = \frac{x^2}{4} + 3x - \frac{3}{x} back into the equation:

dydx=3(x24+3x3x)2(x2+3+3x2)\frac{dy}{dx} = 3 \left( \frac{x^2}{4} + 3x - \frac{3}{x} \right)^2 \cdot \left( \frac{x}{2} + 3 + \frac{3}{x^2} \right)

This is the derivative dydx\frac{dy}{dx} as a function of xx.

Would you like further clarifications or details?

Here are five related questions to further your understanding:

  1. How would the process change if the exponent in the original function was different?
  2. What is the significance of using the chain rule in this context?
  3. Can you identify other problems where expressing functions as y=f(u)y = f(u) and u=g(x)u = g(x) simplifies differentiation?
  4. How would you compute higher-order derivatives for this function?
  5. How can the quotient rule assist in finding dudx\frac{du}{dx} for more complex fractions?

Tip: Always make sure to simplify terms before differentiating to avoid unnecessary complications!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Algebra

Formulas

dy/dx = dy/du * du/dx
d/dx (x^n) = n * x^(n-1)
d/dx (1/x) = -1/x^2

Theorems

Chain Rule

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)