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Let g be a function such that g(9)=0 and g^(')(9)=2. Let h be the function h(x)=sqrtx. Evaluate (d)/(dx)[g(x)*h(x)] at x=9.

- Let g g be a function such that g(9)=0 g(9)=0 and g′(9)=2 g^{\prime}(9)=2 .\newline- Let h h be the function h(x)=x h(x)=\sqrt{x} .\newlineEvaluate ddx[g(x)⋅h(x)] \frac{d}{d x}[g(x) \cdot h(x)] at x=9 x=9 .

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Full solution

Q. - Let g g be a function such that g(9)=0 g(9)=0 and g′(9)=2 g^{\prime}(9)=2 .\newline- Let h h be the function h(x)=x h(x)=\sqrt{x} .\newlineEvaluate ddx[g(x)⋅h(x)] \frac{d}{d x}[g(x) \cdot h(x)] at x=9 x=9 .
  1. Product Rule Explanation: We need to use the product rule for differentiation, which is (f⋅g)′=f′⋅g+f⋅g′(f \cdot g)' = f' \cdot g + f \cdot g'.
  2. Derivative of h(x)h(x): First, let's find the derivative of h(x)=xh(x) = \sqrt{x}. The derivative h′(x)h'(x) is 12x\frac{1}{2\sqrt{x}}.
  3. Calculate h′(9)h'(9): Now we plug in x=9x=9 into h′(x)h'(x) to get h′(9)=129=16h'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}.
  4. Given Values for gg: We know g(9)=0g(9)=0 and g′(9)=2g'(9)=2 from the problem statement.
  5. Product Rule Application: Using the product rule: (g(x)⋅h(x))′(g(x) \cdot h(x))' at x=9x=9 is g′(9)⋅h(9)+g(9)⋅h′(9)g'(9) \cdot h(9) + g(9) \cdot h'(9).
  6. Substitute Values: Substitute the known values: 2×9+0×(16)2 \times \sqrt{9} + 0 \times \left(\frac{1}{6}\right).
  7. Simplify Result: This simplifies to 2×3+0=62 \times 3 + 0 = 6.

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