Math Problem Statement

Find the derivative of the function.

f(x) = (x^3 − 7x^2 + 1)/

x^2

Solution

To find the derivative of the function f(x)=x37x2+1x2f(x) = \frac{x^3 - 7x^2 + 1}{x^2}, we will simplify the function first and then differentiate.

Step 1: Simplify the Function

First, divide each term in the numerator by x2x^2 to simplify the function:

f(x)=x3x27x2x2+1x2f(x) = \frac{x^3}{x^2} - \frac{7x^2}{x^2} + \frac{1}{x^2}

Simplifying each term:

f(x)=x7+x2f(x) = x - 7 + x^{-2}

Step 2: Differentiate Each Term

Now, differentiate each term separately using the power rule, which states that if f(x)=xnf(x) = x^n, then f(x)=nxn1f'(x) = nx^{n-1}.

  1. Differentiate xx: ddx(x)=1\frac{d}{dx}(x) = 1

  2. Differentiate 7-7: ddx(7)=0\frac{d}{dx}(-7) = 0

  3. Differentiate x2x^{-2}: ddx(x2)=2x21=2x3\frac{d}{dx}(x^{-2}) = -2x^{-2-1} = -2x^{-3}

Step 3: Combine the Derivatives

Combine the derivatives of each term to get the derivative of the entire function:

f(x)=1+02x3f'(x) = 1 + 0 - 2x^{-3}

Simplify the expression:

f(x)=12x3f'(x) = 1 - 2x^{-3}

Or, rewriting 2x3-2x^{-3} as 2x3-\frac{2}{x^3}:

f(x)=12x3f'(x) = 1 - \frac{2}{x^3}

Final Answer

The derivative of the function f(x)=x37x2+1x2f(x) = \frac{x^3 - 7x^2 + 1}{x^2} is:

f(x)=12x3f'(x) = 1 - \frac{2}{x^3}

Would you like more details or have any questions?

Here are some related questions:

  1. What is the second derivative of f(x)=x37x2+1x2f(x) = \frac{x^3 - 7x^2 + 1}{x^2}?
  2. How do you find critical points of f(x)=x37x2+1x2f(x) = \frac{x^3 - 7x^2 + 1}{x^2}?
  3. What happens to the derivative as xx approaches 0?
  4. How do you differentiate a quotient using the quotient rule?
  5. What are the inflection points of the function f(x)=x37x2+1x2f(x) = \frac{x^3 - 7x^2 + 1}{x^2}?

Tip: When differentiating functions with fractions, simplifying the expression first can often make the differentiation process easier.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quotient Rule
Power Rule

Formulas

Power rule for differentiation

Theorems

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Suitable Grade Level

Advanced High School / College