Math Problem Statement

Evaluate

𝑑𝑦

𝑑π‘₯

(π‘₯ + 1)(π‘₯ + √π‘₯) at x = 9

Solution

We need to find the derivative of the product (x+1)(x+x)(x + 1)(x + \sqrt{x}) and evaluate it at x=9x = 9. This requires the use of the product rule for differentiation.

Step 1: Apply the Product Rule

The product rule states:

ddx[f(x)β‹…g(x)]=fβ€²(x)β‹…g(x)+f(x)β‹…gβ€²(x)\frac{d}{dx} [f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)

Here, let:

  • f(x)=(x+1)f(x) = (x + 1)
  • g(x)=(x+x)g(x) = (x + \sqrt{x})

Step 2: Find the derivatives

  1. Derivative of f(x)=x+1f(x) = x + 1:

fβ€²(x)=ddx(x+1)=1f'(x) = \frac{d}{dx}(x + 1) = 1

  1. Derivative of g(x)=x+xg(x) = x + \sqrt{x}:

The derivative of xx is 11, and the derivative of x=x1/2\sqrt{x} = x^{1/2} is 12xβˆ’1/2\frac{1}{2} x^{-1/2}.

So,

gβ€²(x)=ddx(x+x)=1+12xβˆ’1/2g'(x) = \frac{d}{dx}(x + \sqrt{x}) = 1 + \frac{1}{2}x^{-1/2}

gβ€²(x)=1+12xg'(x) = 1 + \frac{1}{2\sqrt{x}}

Step 3: Use the Product Rule

Now, apply the product rule:

ddx[(x+1)(x+x)]=fβ€²(x)β‹…g(x)+f(x)β‹…gβ€²(x)\frac{d}{dx} \left[ (x + 1)(x + \sqrt{x}) \right] = f'(x) \cdot g(x) + f(x) \cdot g'(x)

Substitute the values of f(x),fβ€²(x),g(x),f(x), f'(x), g(x), and gβ€²(x)g'(x):

ddx[(x+1)(x+x)]=1β‹…(x+x)+(x+1)β‹…(1+12x)\frac{d}{dx} \left[ (x + 1)(x + \sqrt{x}) \right] = 1 \cdot (x + \sqrt{x}) + (x + 1) \cdot \left( 1 + \frac{1}{2\sqrt{x}} \right)

Step 4: Simplify the expression

Simplify the right-hand side:

=(x+x)+(x+1)(1+12x)= (x + \sqrt{x}) + (x + 1) \left( 1 + \frac{1}{2\sqrt{x}} \right)

Distribute the terms:

=(x+x)+(x+1)+x+12x= (x + \sqrt{x}) + (x + 1) + \frac{x+1}{2\sqrt{x}}

Simplifying further:

=2x+x+1+x+12x= 2x + \sqrt{x} + 1 + \frac{x+1}{2\sqrt{x}}

Step 5: Evaluate at x=9x = 9

Now, substitute x=9x = 9 into the expression:

  1. 9=3\sqrt{9} = 3
  2. x+12x=9+12β‹…3=106=53\frac{x+1}{2\sqrt{x}} = \frac{9+1}{2\cdot 3} = \frac{10}{6} = \frac{5}{3}

So, the expression becomes:

2(9)+3+1+532(9) + 3 + 1 + \frac{5}{3}

Simplify:

=18+3+1+53= 18 + 3 + 1 + \frac{5}{3}

Convert the whole numbers to fractions with a denominator of 3:

=18β‹…33+3β‹…33+1β‹…33+53= \frac{18 \cdot 3}{3} + \frac{3 \cdot 3}{3} + \frac{1 \cdot 3}{3} + \frac{5}{3} =543+93+33+53= \frac{54}{3} + \frac{9}{3} + \frac{3}{3} + \frac{5}{3} =713= \frac{71}{3}

Conclusion:

The derivative dydx\frac{dy}{dx} of (x+1)(x+x)(x + 1)(x + \sqrt{x}) at x=9x = 9 is 713\frac{71}{3}, or approximately 23.6723.67.


Would you like more explanation on any part of the process?

Related Questions:

  1. What is the product rule and how can it be applied to more complex functions?
  2. How do you differentiate products of functions involving radicals?
  3. How can you apply the chain rule in cases involving x\sqrt{x}?
  4. What other methods exist for finding derivatives when multiple terms are involved?
  5. How do you evaluate derivatives for higher-order terms?

Tip:

When differentiating products of functions, always apply the product rule carefully and simplify the expressions step by step to avoid errors!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Product Rule

Formulas

Product Rule: (f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x)
Derivative of √x: (x^(1/2))' = (1/2) * x^(-1/2)

Theorems

Product Rule

Suitable Grade Level

Grades 11-12