Determine h^(')(1) if h(x)=f(sqrt(g(2-x^(2))):} if g(1)=4,f^(')(2)=2, g(1)=(1)/(4),g^(')(1)=(1)/(2)

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Determine  h^(')(1) if  h(x)=f(sqrt(g(2-x^(2))):} if  g(1)=4,f^(')(2)=2,  g(1)=(1)/(4),g^(')(1)=(1)/(2)

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Find Derivative at x=1: We need to find the derivative of h(x) at x = 1. To do this, we will use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. The function h(x) is a composition of f and √(g(2 - x²)), so we will need to differentiate both of these functions and then evaluate at x = 1.Use Chain Rule: First, let's find the derivative of the inner function √(g(2 - x²)). We will use the chain rule again, this time for g(2 - x²) and the square root function. The derivative of √u with respect to u is (1/2)u^(-1/2), and the derivative of g(2 - x²) with respect to x is -2xg'(2 - x²) by the chain rule.Derivative of Inner Function: Now we can write the derivative of the inner function as: (1/2)(g(2 - x²))^(-1/2) * (-2xg'(2 - x²)).Derivative of Outer Function: Next, we need to find the derivative of the outer function f(u) with respect to u, which is simply f'(u). We will evaluate this derivative at u = √(g(2 - x²)).Evaluate at x=1: Now we can write the derivative of h(x) as: h'(x) = f'(√(g(2 - x²))) * (1/2)(g(2 - x²))^(-1/2) * (-2xg'(2 - x²)).Calculate g(1): To find h'(1), we need to evaluate this expression at x = 1. We will need the values of g(1), g'(1), and f'(2) to do this. We are given that g(1) = 4, g'(1) = 1/2, and f'(2) = 2.Evaluate f'(2): First, we calculate g(2 - 1²) = g(1) = 4. Then we find the square root of this value, which is √4 = 2.Evaluate g'(1): Next, we evaluate f'(√(g(2 - 1²))) = f'(2) = 2, since we are given that f'(2) = 2.Substitute Values: We also need to evaluate g'(2 - 1²) = g'(1) = 1/2, as given.Simplify Expression: Now we can substitute all the values into the derivative of h(x) to find h'(1): h'(1) = 2 * (1/2)(4)^(-1/2) * (-2*1*(1/2)).Simplifying the expression, we get: h'(1) = 2 * (1/2)(1/2) * (-2*1*(1/2)) = 2 * (1/4) * (-1) = -1/2.

Determine $h'(1)$ if $h(x)=f(\sqrt{g(2-x^2)})$ : if $g(1)=4$ , $f'(2)=2$ , $g(1)=\frac{1}{4}$ , $g'(1)=\frac{1}{2}$

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Determine h′(1)h'(1)h′(1) if h(x)=f(g(2−x2))h(x)=f(\sqrt{g(2-x^2)})h(x)=f(g(2−x2)​): if g(1)=4g(1)=4g(1)=4, f′(2)=2f'(2)=2f′(2)=2, g(1)=14g(1)=\frac{1}{4}g(1)=41​, g′(1)=12g'(1)=\frac{1}{2}g′(1)=21​

Determine $h'(1)$ if $h(x)=f(\sqrt{g(2-x^2)})$ : if $g(1)=4$ , $f'(2)=2$ , $g(1)=\frac{1}{4}$ , $g'(1)=\frac{1}{2}$