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Let h(x)=7-x-2x^(5) and let f be the inverse function of h. Notice that h(-1)=10. f^(')(10)=

Let h(x)=7−x−2x5 h(x)=7-x-2 x^{5} and let f f be the inverse function of h h . Notice that h(−1)=10 h(-1)=10 .\newlinef′(10)= f^{\prime}(10)=

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Full solution

Q. Let h(x)=7−x−2x5 h(x)=7-x-2 x^{5} and let f f be the inverse function of h h . Notice that h(−1)=10 h(-1)=10 .\newlinef′(10)= f^{\prime}(10)=
  1. Use Inverse Function Derivative Formula: To find f′(10)f^{\prime}(10), we need to use the formula for the derivative of the inverse function: f′(x)=1h′(f(x)).f^{\prime}(x) = \frac{1}{h^{\prime}(f(x))}.
  2. Find Derivative of h(x)h(x): First, we need to find h′(x)h^{\prime}(x) which is the derivative of h(x)=7−x−2x5h(x) = 7 - x - 2x^{5}.\newlineh′(x)=−1−10x4h^{\prime}(x) = -1 - 10x^{4}.
  3. Calculate h'(\-1): Since h(\-1) = 10, we need to find h'(\-1) to use in our formula.\newline$h'((-1\)) = (-1\) - \(10\)((-1\))^{\(4\)} = (-1\) - \(10\)(\(1\)) = (-1\) - \(10\) = (-11\).
  4. Find \(f'(10)\): Now we can find \(f'(10)\) using the formula.\(\newline\)\(f'(10) = \frac{1}{h'(f(10))} = \frac{1}{h'(-1)} = \frac{1}{-11}.\)
  5. Final Result: So, \(f^{\prime}(10) = -\frac{1}{11}.\)

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