Math Problem Statement

2π‘₯ 2 +5π‘₯+3 π‘₯+9 diff wrt x

Solution

To differentiate the expression 2x2+5x+3x+9\frac{2x^2 + 5x + 3}{x + 9} with respect to xx, we will use the quotient rule. The quotient rule states that if we have a function f(x)=u(x)v(x)f(x) = \frac{u(x)}{v(x)}, then its derivative is given by:

fβ€²(x)=v(x)β‹…uβ€²(x)βˆ’u(x)β‹…vβ€²(x)(v(x))2f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}

Here, we have:

  • u(x)=2x2+5x+3u(x) = 2x^2 + 5x + 3
  • v(x)=x+9v(x) = x + 9

Now, let's differentiate u(x)u(x) and v(x)v(x):

  • uβ€²(x)=4x+5u'(x) = 4x + 5
  • vβ€²(x)=1v'(x) = 1

Using the quotient rule:

fβ€²(x)=(x+9)β‹…(4x+5)βˆ’(2x2+5x+3)β‹…1(x+9)2f'(x) = \frac{(x + 9) \cdot (4x + 5) - (2x^2 + 5x + 3) \cdot 1}{(x + 9)^2}

Now we expand and simplify:

fβ€²(x)=(x+9)(4x+5)βˆ’(2x2+5x+3)(x+9)2f'(x) = \frac{(x + 9)(4x + 5) - (2x^2 + 5x + 3)}{(x + 9)^2}

Expanding both terms:

(x+9)(4x+5)=4x2+5x+36x+45=4x2+41x+45(x + 9)(4x + 5) = 4x^2 + 5x + 36x + 45 = 4x^2 + 41x + 45

Now subtract (2x2+5x+3)(2x^2 + 5x + 3):

fβ€²(x)=(4x2+41x+45)βˆ’(2x2+5x+3)(x+9)2f'(x) = \frac{(4x^2 + 41x + 45) - (2x^2 + 5x + 3)}{(x + 9)^2}

Simplifying the numerator:

fβ€²(x)=2x2+36x+42(x+9)2f'(x) = \frac{2x^2 + 36x + 42}{(x + 9)^2}

So, the derivative of the function is:

fβ€²(x)=2x2+36x+42(x+9)2f'(x) = \frac{2x^2 + 36x + 42}{(x + 9)^2}

Would you like more details or further clarifications?

Here are 5 related questions:

  1. How is the quotient rule applied to more complex functions?
  2. What other methods can be used to differentiate rational functions?
  3. Can this differentiation be solved using implicit differentiation?
  4. How do higher-order derivatives of rational functions behave?
  5. What is the significance of critical points in such rational functions?

Tip: The quotient rule is particularly useful for differentiating rational functions where both the numerator and denominator involve xx. Always apply it carefully to avoid mistakes in expanding terms.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Quotient Rule

Formulas

f'(x) = (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2
Derivative of u(x) = 4x + 5
Derivative of v(x) = 1

Theorems

Quotient Rule for Differentiation

Suitable Grade Level

Undergraduate Calculus