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Given that f(x) = 2x + cos(x) is one-to-one, use the formula (f^(-1))^'(x) = 1 / f'(f^(-1)(x)) to find (f^(-1))^'(1) = ?

User Toren
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Final answer:

To find (f^(-1))^'(1), we need to find f^(-1)(1) and f'(f^(-1)(1)). Using the formula (f^(-1))^'(x) = 1 / f'(f^(-1)(x)), we can substitute these values to find the derivative of the inverse function.

Step-by-step explanation:

To find the derivative of the inverse function, we can use the formula (f^(-1))^'(x) = 1 / f'(f^(-1)(x)). In this case, we want to find (f^(-1))^'(1). So we need to find f^(-1)(1) and f'(f^(-1)(1)).

First, we find f^(-1)(1) by setting 2x + cos(x) = 1 and solving for x. This gives us x = f^(-1)(1) = 0.450183.

Next, we find f'(f^(-1)(1)). The derivative of f(x) = 2x + cos(x) is f'(x) = 2 - sin(x). Plugging in x = f^(-1)(1) = 0.450183, we get f'(f^(-1)(1)) = 2 - sin(0.450183) = 1.71634.

Finally, we can substitute these values into the formula (f^(-1))^'(x) = 1 / f'(f^(-1)(x)) to find (f^(-1))^'(1) = 1 / f'(f^(-1)(1)) = 1 / 1.71634 = 0.582845.'

User Jishnu Raj T
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