6. (2 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 \( n D e r i v \).
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To find \( \left.\frac{d f}{d x}\right|_{x=2} \), we first need to compute \( k(x) \) and then find \( f(k) \). Since \( f(k) = 7 - 3 \ln(k) \), we'll use the chain rule for differentiation: \[ \frac{d f}{d x} = \frac{d f}{d k} \cdot \frac{d k}{d x} \] First, calculate \( k(2) \): \[ k(2) = 1.234(2^4) + 4.56(2^2) + 7.89 = 1.234(16) + 4.56(4) + 7.89 \] \[ = 19.744 + 18.24 + 7.89 = 45.874 \] Next, we find \( f(k(2)) \): \[ f(k(2)) = 7 - 3 \ln(45.874) \] Now, calculate \( \frac{d f}{d k} \): \[ \frac{d f}{d k} = -\frac{3}{k} \] For \( \frac{d k}{d x} \), find the derivative of \( k(x) \): \[ k(x) = 1.234 x^4 + 4.56 x^2 + 7.89 \] \[ \frac{d k}{d x} = 4(1.234) x^3 + 2(4.56) x = 4.936 x^3 + 9.12 x \] Evaluate at \( x = 2 \): \[ \frac{d k}{d x}\bigg|_{x=2} = 4.936(2^3) + 9.12(2) = 4.936(8) + 18.24 = 39.488 + 18.24 = 57.728 \] Now, combine these: \[ \frac{d f}{d x} = -\frac{3}{k(2)} \cdot \frac{d k}{d x} \] \[ = -\frac{3}{45.874} \cdot 57.728 \approx -3.773 \] Thus, \( \left.\frac{d f}{d x}\right|_{x=2} \approx -3.773 \).
