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The functions c(x) c(x) and d(x) d(x) are differentiable. The function r(x) r(x) is defined as: r(x)=c(x)d(x) r(x)= \frac{c(x)}{d(x)} If c(6)=7 c(6)= 7 , c′(6)=5 c'(6)= 5 , d(6)=3 d(6)= 3 , and d′(6)=−2 d'(6)= -2 , what is r′(6) r'(6) ? Simplify any fractions. r′(6)= r'(6)=

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Q. The functions c(x) c(x) and d(x) d(x) are differentiable. The function r(x) r(x) is defined as: r(x)=c(x)d(x) r(x)= \frac{c(x)}{d(x)} If c(6)=7 c(6)= 7 , c′(6)=5 c'(6)= 5 , d(6)=3 d(6)= 3 , and d′(6)=−2 d'(6)= -2 , what is r′(6) r'(6) ? Simplify any fractions. r′(6)= r'(6)=
  1. Given function: We are given the function r(x)=c(x)d(x) r(x) = \frac{c(x)}{d(x)} and we need to find the derivative of r r at x=6 x = 6 , denoted as r′(6) r'(6) . To do this, we will use the quotient rule for derivatives, which states that if h(x)=f(x)g(x) h(x) = \frac{f(x)}{g(x)} , then h′(x)=f′(x)g(x)−f(x)g′(x)(g(x))2 h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} .
  2. Applying quotient rule: We have c(6)=7c(6) = 7, c′(6)=5c'(6) = 5, d(6)=3d(6) = 3, and d′(6)=−2d'(6) = -2. Let's apply the quotient rule using these values:\newliner′(6)=c′(6)d(6)−c(6)d′(6)(d(6))2.r'(6) = \frac{c'(6)d(6) - c(6)d'(6)}{(d(6))^2}.
  3. Substituting values: Substitute the given values into the quotient rule formula: r′(6)=(5⋅3−7⋅−2)(3)2 r'(6) = \frac{(5 \cdot 3 - 7 \cdot -2)}{(3)^2} .
  4. Performing multiplication: Perform the multiplication: r′(6)=15+149r'(6) = \frac{15 + 14}{9}.
  5. Adding numerator: Add the numbers in the numerator: r′(6)=299r'(6) = \frac{29}{9}.
  6. Final answer: The fraction 299\frac{29}{9} is already simplified, so this is our final answer.

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