Find the derivative of the function. \[ y=\frac{5}{e^{6 x}}+\frac{e^{8 x}}{2} \] \( 30 e^{6 x}+16 x e^{8 x 2} \)

We are given the function \[ y=\frac{5}{e^{6x}}+\frac{e^{8x}}{2}. \] **Step 1. Rewrite the function** We can express the function using negative exponents for a simpler differentiation: \[ y=5e^{-6x}+\frac{1}{2}e^{8x}. \] **Step 2. Differentiate each term** Differentiate term-by-term. - For the first term, \( 5e^{-6x} \): \[ \frac{d}{dx}\left(5e^{-6x}\right)=5\cdot \frac{d}{dx}\left(e^{-6x}\right). \] We use the chain rule where the derivative of \( e^{u} \) is \( e^{u}\,\frac{du}{dx} \). Here, \( u=-6x \) and \( \frac{du}{dx}=-6 \). Thus, \[ \frac{d}{dx}\left(e^{-6x}\right)=e^{-6x}\cdot (-6)=-6e^{-6x}. \] Therefore, \[ \frac{d}{dx}\left(5e^{-6x}\right)=5\cdot \left(-6e^{-6x}\right)=-30e^{-6x}. \] - For the second term, \( \frac{1}{2}e^{8x} \): \[ \frac{d}{dx}\left(\frac{1}{2}e^{8x}\right)=\frac{1}{2}\cdot\frac{d}{dx}\left(e^{8x}\right). \] Again, applying the chain rule with \( u=8x \) (so \( \frac{du}{dx}=8 \)) yields \[ \frac{d}{dx}\left(e^{8x}\right)=e^{8x}\cdot 8=8e^{8x}. \] Thus, \[ \frac{d}{dx}\left(\frac{1}{2}e^{8x}\right)=\frac{1}{2}\cdot 8e^{8x}=4e^{8x}. \] **Step 3. Combine the results** The derivative of the function is the sum of the derivatives: \[ y'=-30e^{-6x}+4e^{8x}. \] Thus, the final answer is: \[ y'=-30e^{-6x}+4e^{8x}. \]