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The functions s(x) s(x) and t(x) t(x) are differentiable. \newlineThe function u(x) u(x) is defined as: u(x)=s(x)t(x) u(x)= \frac{s(x)}{t(x)} \newlineIf s(6)=7 s(6)= 7 , s′(6)=5 s'(6)= 5 , t(6)=9 t(6)= 9 , and t′(6)=2 t'(6)= 2 , what is u′(6) u'(6) ? \newlineSimplify any fractions. \newlineu′(6)= u'(6)= _____

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Q. The functions s(x) s(x) and t(x) t(x) are differentiable. \newlineThe function u(x) u(x) is defined as: u(x)=s(x)t(x) u(x)= \frac{s(x)}{t(x)} \newlineIf s(6)=7 s(6)= 7 , s′(6)=5 s'(6)= 5 , t(6)=9 t(6)= 9 , and t′(6)=2 t'(6)= 2 , what is u′(6) u'(6) ? \newlineSimplify any fractions. \newlineu′(6)= u'(6)= _____
  1. Quotient Rule for Derivatives: To find u′(6)u'(6), we need to use the quotient rule for derivatives, which states that if u(x)=s(x)t(x)u(x) = \frac{s(x)}{t(x)}, then u′(x)=s′(x)t(x)−s(x)t′(x)(t(x))2u'(x) = \frac{s'(x)t(x) - s(x)t'(x)}{(t(x))^2}. We will apply this rule using the given values.
  2. Calculating the Numerator: First, we calculate the numerator of the quotient rule: s′(6)t(6)−s(6)t′(6)=5×9−7×2s'(6)t(6) - s(6)t'(6) = 5 \times 9 - 7 \times 2.
  3. Numerator Calculation: Performing the calculation for the numerator: 5×9−7×2=45−14=315 \times 9 - 7 \times 2 = 45 - 14 = 31.
  4. Calculating the Denominator: Next, we calculate the denominator of the quotient rule: (t(6))2=92(t(6))^2 = 9^2.
  5. Denominator Calculation: Performing the calculation for the denominator: 92=819^2 = 81.
  6. Finding u′(6)u'(6): Now, we can find u′(6)u'(6) by dividing the numerator by the denominator: u′(6)=3181u'(6) = \frac{31}{81}.
  7. Conclusion: We have found the derivative of uu at x=6x = 6, so we can conclude the solution.

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