Questions: Find (k'(x)) if (k(x)=-frac5x^3 cdot e^-3 x^4+4 x^2) Provide your answer below: [ k^prime(x)= ]

$Find (k'(x)) if (k(x)=-frac5x^3 cdot e^-3 x^4+4 x^2) Provide your answer below: [ k^prime(x)= ]$

Transcript text: Find $k^{\prime}(x)$ if $k(x)=-\frac{5}{x^{3}} \cdot e^{-3 x^{4}+4 x^{2}}$ Provide your answer below: \[ k^{\prime}(x)= \]

Solution

Solution Steps

To find the derivative of the function $ k(x) = -\frac{5}{x^3} \cdot e^{-3x^4 + 4x^2} $, we will use the product rule and the chain rule. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. The chain rule is used to differentiate composite functions. We will apply these rules to find $ k'(x) $.

Step 1: Define the Function

We start with the function given by \[ k(x) = -\frac{5}{x^3} \cdot e^{-3x^4 + 4x^2}. \]

Step 2: Apply the Product Rule

To find the derivative $ k'(x) $, we apply the product rule: \[ k'(x) = u'v + uv', \] where $ u = -\frac{5}{x^3} $ and $ v = e^{-3x^4 + 4x^2} $.

Step 3: Calculate Derivatives

We compute the derivatives:

For $ u $: \[ u' = \frac{d}{dx}\left(-\frac{5}{x^3}\right) = 15x^{-4} = \frac{15}{x^4}. \]
For $ v $: Using the chain rule, \[ v' = e^{-3x^4 + 4x^2} \cdot \frac{d}{dx}(-3x^4 + 4x^2) = e^{-3x^4 + 4x^2} \cdot (-12x^3 + 8x). \]

Step 4: Substitute Back into the Product Rule

Substituting $ u, u', v, $ and $ v' $ back into the product rule gives: \[ k'(x) = \left(\frac{15}{x^4}\right) e^{-3x^4 + 4x^2} + \left(-\frac{5}{x^3}\right) \left(e^{-3x^4 + 4x^2} \cdot (-12x^3 + 8x)\right). \]

Step 5: Simplify the Expression

Combining the terms, we have: \[ k'(x) = \frac{15 e^{-3x^4 + 4x^2}}{x^4} + \frac{5(-12x^3 + 8x)e^{-3x^4 + 4x^2}}{x^3}. \] This simplifies to: \[ k'(x) = e^{-3x^4 + 4x^2} \left(\frac{15}{x^4} + \frac{5(-12x^3 + 8x)}{x^3}\right). \]

Step 6: Final Simplification

Further simplifying the expression inside the parentheses: \[ k'(x) = e^{-3x^4 + 4x^2} \left(\frac{15}{x^4} - \frac{60x^3}{x^3} + \frac{40x}{x^3}\right) = e^{-3x^4 + 4x^2} \left(\frac{15 - 60x^4 + 40x^2}{x^4}\right). \]