6. \( \left(2\right. \) pts) If \( k(x)=1.234 x^{4}+4.56 x^{2}+7.89 \) and \( f(k)=7-3 \ln k \), find \( \left.\frac{d f}{d x}\right|_{x=2} \) Use your calculator \( Y= \) list and \( n \) Deriv.

To find \( \left.\frac{d f}{d x}\right|_{x=2} \), we need to apply the chain rule to differentiate \( f(k(x)) \) with respect to \( x \). Start with the function \( f(k) = 7 - 3 \ln(k) \). We know that the derivative of \( f \) with respect to \( k \) is: \[ \frac{df}{dk} = -\frac{3}{k} \] Next, we differentiate \( k(x) = 1.234 x^4 + 4.56 x^2 + 7.89 \): \[ \frac{dk}{dx} = 4 \cdot 1.234 x^3 + 2 \cdot 4.56 x = 4.936 x^3 + 9.12 x \] Using the chain rule, we have: \[ \frac{df}{dx} = \frac{df}{dk} \cdot \frac{dk}{dx} \] Substituting in the values we calculated: \[ \frac{df}{dx} = -\frac{3}{k} \cdot (4.936 x^3 + 9.12 x) \] Now we need to evaluate this expression at \( x = 2 \): 1. Calculate \( k(2) \): \[ k(2) = 1.234(2^4) + 4.56(2^2) + 7.89 = 1.234(16) + 4.56(4) + 7.89 = 19.744 + 18.24 + 7.89 = 45.874 \] 2. Determine \( \frac{dk}{dx} \) at \( x = 2 \): \[ \frac{dk}{dx} = 4.936(2^3) + 9.12(2) = 4.936(8) + 9.12(2) = 39.488 + 18.24 = 57.728 \] 3. Now plug everything back into the derivative: \[ \frac{df}{dx} \bigg|_{x=2} = -\frac{3}{45.874} \cdot 57.728 \] Calculating this: \[ \frac{df}{dx} \bigg|_{x=2} \approx -\frac{173.184}{45.874} \approx -3.77 \] So, \( \left.\frac{d f}{d x}\right|_{x=2} \approx -3.77 \).