**Problem:**Find \( y'' \) for \( y = e^{x^3} + 11x \).**Solution:**First, find the first derivative \( y' \).To differentiate \( y = e^{x^3} \), use the chain rule:\[ \frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2. \]For \( 11x \), the derivative is:\[ \frac{d}{dx}(11x) = 11. \]So,\[ y' = 3x^2 e^{x^3} + 11. \]Next, find the second derivative \( y'' \).Differentiate \( y' = 3x^2 e^{x^3} + 11 \):\[ y'' = \frac{d}{dx}(3x^2 e^{x^3}) + \frac{d}{dx}(11). \]Differentiate \( 3x^2 e^{x^3} \) using the product rule (\( u'v + uv' \)), where \( u = 3x^2 \) and \( v = e^{x^3} \).- \( u' = \frac{d}{dx}(3x^2) = 6x \)- \( v' = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2 = 3x^2 e^{x^3} \)So,\[ \frac{d}{dx}(3x^2 e^{x^3}) = (6x)(e^{x^3}) + (3x^2)(3x^2 e^{x^3}) = 6x e^{x^3} + 9x^4 e^{x^3}. \]Since the derivative of \( 11 \) is \( 0 \), we have:\[ y'' = 6x e^{x^3} + 9x^4 e^{x^3}. \]Therefore, the second derivative is:\[ y'' = (6x + 9x^4)e^{x^

Name: **Problem:**Find \( y'' \) for \( y = e^{x^3} + 11x \).**Solution:**F
Uploaded: 2024-03-20T06:47:22.000Z
Channel: Bartleby
Description: **Problem:**Find \( y'' \) for \( y = e^{x^3} + 11x \).**Solution:**First, find the first derivative \( y' \).To differentiate \( y = e^{x^3} \), use the chain rule:\[ \frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2. \]F