Math Problem Statement

If f(x) = \frac{(x-1)(2x-1)}{\sqrt{5x+4}}, then what is f'(1)?

Solution

To determine f(1)f'(1) when f(x)=(x1)(2x1)5x+4f(x) = \frac{(x-1)(2x-1)}{\sqrt{5x+4}}, we use differentiation rules and evaluate at x=1x = 1.


1. Function Setup

The given function is:

f(x)=(x1)(2x1)5x+4f(x) = \frac{(x-1)(2x-1)}{\sqrt{5x+4}}

This is a quotient of two functions, so we need the quotient rule for derivatives.


2. Derivative Using Quotient Rule

The quotient rule states:

(uv)=uvuvv2\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}

Here:

  • u(x)=(x1)(2x1)u(x) = (x-1)(2x-1)
  • v(x)=5x+4v(x) = \sqrt{5x+4}

Step 1: Differentiate u(x)u(x)

We simplify u(x)u(x) first:

u(x)=(x1)(2x1)    u(x)=2x23x+1u(x) = (x-1)(2x-1) \implies u(x) = 2x^2 - 3x + 1

Now differentiate:

u(x)=ddx[2x23x+1]=4x3u'(x) = \frac{d}{dx}[2x^2 - 3x + 1] = 4x - 3

Step 2: Differentiate v(x)v(x)

The denominator v(x)=5x+4v(x) = \sqrt{5x+4}. Using the chain rule:

v(x)=ddx((5x+4)12)=12(5x+4)125=525x+4v'(x) = \frac{d}{dx} \left( (5x+4)^{\frac{1}{2}} \right) = \frac{1}{2}(5x+4)^{-\frac{1}{2}} \cdot 5 = \frac{5}{2\sqrt{5x+4}}


3. Apply Quotient Rule

Substitute into the quotient rule:

f(x)=u(x)v(x)u(x)v(x)v(x)2f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}

  1. u(x)=4x3u'(x) = 4x - 3
  2. v(x)=5x+4v(x) = \sqrt{5x+4}
  3. v(x)=525x+4v'(x) = \frac{5}{2\sqrt{5x+4}}

At x=1x = 1, we calculate the following:

Step 1: Compute u(1)u(1) and u(1)u'(1)

  • u(x)=2x23x+1u(1)=2(1)23(1)+1=0u(x) = 2x^2 - 3x + 1 \Rightarrow u(1) = 2(1)^2 - 3(1) + 1 = 0
  • u(x)=4x3u(1)=4(1)3=1u'(x) = 4x - 3 \Rightarrow u'(1) = 4(1) - 3 = 1

Step 2: Compute v(1)v(1) and v(1)v'(1)

  • v(x)=5x+4v(1)=5(1)+4=9=3v(x) = \sqrt{5x+4} \Rightarrow v(1) = \sqrt{5(1) + 4} = \sqrt{9} = 3
  • v(x)=525x+4v(1)=529=56v'(x) = \frac{5}{2\sqrt{5x+4}} \Rightarrow v'(1) = \frac{5}{2\sqrt{9}} = \frac{5}{6}

Step 3: Substitute into the Quotient Rule

Now:

f(1)=u(1)v(1)u(1)v(1)v(1)2f'(1) = \frac{u'(1)v(1) - u(1)v'(1)}{v(1)^2}

Substitute the values:

  • u(1)=1u'(1) = 1, u(1)=0u(1) = 0, v(1)=3v(1) = 3, v(1)=56v'(1) = \frac{5}{6}

f(1)=(1)(3)(0)(56)32f'(1) = \frac{(1)(3) - (0)\left( \frac{5}{6} \right)}{3^2}

Simplify:

f(1)=309=39=13f'(1) = \frac{3 - 0}{9} = \frac{3}{9} = \frac{1}{3}


4. Final Answer

The correct answer is:

13(پ)\boxed{\frac{1}{3} \, \text{(پ)}}


Do you want further details or explanations on any step?


Related Questions:

  1. What is the derivative of a product of two functions (product rule)?
  2. How do you differentiate a square root function using the chain rule?
  3. How does the quotient rule work for derivatives?
  4. What happens if a numerator or denominator equals zero at a specific point?
  5. How do you simplify derivatives involving radicals?

Tip:

Always simplify expressions as much as possible before differentiating to make calculations easier.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quotient Rule
Chain Rule

Formulas

Quotient Rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)
Chain Rule: \( \frac{d}{dx} [g(h(x))] = g'(h(x)) \cdot h'(x) \)

Theorems

Quotient Rule for Derivatives
Chain Rule for Derivatives

Suitable Grade Level

Undergraduate Calculus, AP Calculus AB/BC