Calc I: Find the dervative: (4) $$ y=(4 x+5)\left(\frac{3 x}{2 x^{2}+8}\right) $$

Question

UpStudy AI · Accepted Answer

Let 
$$
y = (4x+5) \cdot \frac{3x}{2x^2+8}.
$$

We will use the product rule, which states
$$
y' = u'(x)v(x) + u(x)v'(x),
$$
with 
$$
u(x) = 4x+5 \quad \text{and} \quad v(x) = \frac{3x}{2x^2+8}.
$$

1. **Differentiate $$u(x)$$:**  
   $$
   u'(x) = \frac{d}{dx}(4x+5) = 4.
   $$

2. **Differentiate $$v(x)$$ using the quotient rule:**  
   The quotient rule for $$ v(x) = \frac{f(x)}{g(x)} $$ is
   $$
   v'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.
   $$
   Here,
   $$
   f(x) = 3x \quad \text{and} \quad g(x) = 2x^2+8.
   $$
   - Differentiate $$f(x)$$:
     $$
     f'(x) = 3.
     $$
   - Differentiate $$g(x)$$:
     $$
     g'(x) = \frac{d}{dx}(2x^2+8) = 4x.
     $$
   Now, apply the quotient rule:
   $$
   v'(x) = \frac{3(2x^2+8) - 3x(4x)}{(2x^2+8)^2}.
   $$
   Simplify the numerator:
   $$
   3(2x^2+8) = 6x^2+24,\quad \text{and} \quad 3x(4x) = 12x^2.
   $$
   So,
   $$
   v'(x) = \frac{6x^2+24 - 12x^2}{(2x^2+8)^2} = \frac{-6x^2+24}{(2x^2+8)^2}.
   $$
   Factor the numerator:
   $$
   -6x^2+24 = 6(4-x^2),
   $$
   hence,
   $$
   v'(x) = \frac{6(4-x^2)}{(2x^2+8)^2}.
   $$

3. **Apply the product rule:**  
   $$
   y' = u'(x)v(x) + u(x)v'(x) = 4 \cdot \frac{3x}{2x^2+8} + (4x+5) \cdot \frac{6(4-x^2)}{(2x^2+8)^2}.
   $$
   Thus, the derivative is:
   $$
   y' = \frac{12x}{2x^2+8} + \frac{6(4-x^2)(4x+5)}{(2x^2+8)^2}.
   $$
The derivative of $$ y = (4x+5) \cdot \frac{3x}{2x^2+8} $$ is: $$ y' = \frac{12x}{2x^2+8} + \frac{6(4-x^2)(4x+5)}{(2x^2+8)^2}. $$

Calc I: Find the dervative: (4) \( y=(4 x+5)\left(\frac{3 x}{2 x^{2}+8}\right) \)

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions