Math Problem Statement

{x^5}{(x^2+4)^2} find this Different fractions

Solution

To find the derivative of the expression:

y=x5(x2+4)2y = \frac{x^5}{(x^2 + 4)^2}

we'll use the quotient rule along with the product rule and chain rule as needed.

The quotient rule for differentiation states that:

ddx(f(x)g(x))=f(x)g(x)f(x)g(x)(g(x))2\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2}

Here:

  • f(x)=x5f(x) = x^5 and
  • g(x)=(x2+4)2g(x) = (x^2 + 4)^2

Step 1: Differentiate f(x)=x5f(x) = x^5

f(x)=5x4f'(x) = 5x^4

Step 2: Differentiate g(x)=(x2+4)2g(x) = (x^2 + 4)^2 using the chain rule

Let u(x)=x2+4u(x) = x^2 + 4, so g(x)=u2g(x) = u^2.

Now, differentiate u(x)=x2+4u(x) = x^2 + 4: u(x)=2xu'(x) = 2x

Now apply the chain rule to differentiate g(x)=u2g(x) = u^2: g(x)=2uu(x)=2(x2+4)2x=4x(x2+4)g'(x) = 2u \cdot u'(x) = 2(x^2 + 4) \cdot 2x = 4x(x^2 + 4)

Step 3: Apply the quotient rule

Now, substitute into the quotient rule formula: dydx=(5x4)(x2+4)2(x5)(4x(x2+4))(x2+4)4\frac{dy}{dx} = \frac{(5x^4)(x^2 + 4)^2 - (x^5)(4x(x^2 + 4))}{(x^2 + 4)^4}

Step 4: Simplify the expression

Simplify the numerator: Numerator=5x4(x2+4)24x6(x2+4)\text{Numerator} = 5x^4(x^2 + 4)^2 - 4x^6(x^2 + 4)

We can factor out x4(x2+4)x^4(x^2 + 4) from both terms in the numerator: Numerator=x4(x2+4)[5(x2+4)4x2]\text{Numerator} = x^4(x^2 + 4) \left[ 5(x^2 + 4) - 4x^2 \right]

Simplify the terms inside the brackets: 5(x2+4)4x2=5x2+204x2=x2+205(x^2 + 4) - 4x^2 = 5x^2 + 20 - 4x^2 = x^2 + 20

So, the numerator becomes: Numerator=x4(x2+4)(x2+20)\text{Numerator} = x^4(x^2 + 4)(x^2 + 20)

Now, the derivative is: dydx=x4(x2+4)(x2+20)(x2+4)4\frac{dy}{dx} = \frac{x^4(x^2 + 4)(x^2 + 20)}{(x^2 + 4)^4}

Finally, cancel one factor of (x2+4)(x^2 + 4) from the numerator and the denominator: dydx=x4(x2+20)(x2+4)3\frac{dy}{dx} = \frac{x^4(x^2 + 20)}{(x^2 + 4)^3}

This is the simplified derivative.

Would you like more details, or have any questions?

Related Questions:

  1. How do you apply the product rule in differentiation?
  2. Can you explain the chain rule in more detail?
  3. How does implicit differentiation work in calculus?
  4. What are some common mistakes when using the quotient rule?
  5. How does the power rule for differentiation work?

Tip:

When differentiating complex expressions, always simplify as much as possible before applying rules like the quotient rule—it helps avoid unnecessary complications.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quotient Rule
Product Rule
Chain Rule

Formulas

Quotient rule: d/dx [f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2
Chain rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)
Product rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

Theorems

Quotient rule
Chain rule
Product rule

Suitable Grade Level

Undergraduate Calculus