Let f be a function such that f(-2)=8 and f^(')(-2)=4. Let h be the function h(x)=x^(3). Evaluate (d)/(dx)[(f(x))/(h(x))] at x=-2.

Question

Let  f be a function such that  f(-2)=8 and  f^(')(-2)=4. Let  h be the function  h(x)=x^(3).  Evaluate  (d)/(dx)[(f(x))/(h(x))] at  x=-2.

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Given functions and rule: We are given the function f(x) and its derivative at x=-2, as well as the function h(x). To find the derivative of the quotient (f(x)/h(x)), we will use the quotient rule, which states that the derivative of a quotient is given by: (d/dx)[(f(x)/h(x))] = (h(x)f'(x) - f(x)h'(x)) / (h(x))^2Find derivative of h(x): First, we need to find the derivative of h(x), which is h'(x). Since h(x) = x^3, we can use the power rule to find h'(x). The power rule states that the derivative of x^n is n*x^(n-1). So, h'(x) = d/dx[x^3] = 3*x^2.Evaluate h'(x) at x=-2: Now we need to evaluate h'(x) at x=-2. Plugging in -2 for x, we get: h'(-2) = 3*(-2)^2 = 3*4 = 12.Evaluate h(x) at x=-2: Next, we need to evaluate h(x) at x=-2. Plugging in -2 for x in h(x) = x^3, we get: h(-2) = (-2)^3 = -8.Apply quotient rule: Now we have all the values we need to apply the quotient rule: f(-2) = 8 (given) f'(-2) = 4 (given) h(-2) = -8 (calculated) h'(-2) = 12 (calculated)  Using the quotient rule: (d/dx)[(f(x)/h(x))] at x=-2 = (h(-2)f'(-2) - f(-2)h'(-2)) / (h(-2))^2Calculate f(-2), f'(-2), h(-2), h'(-2): Plugging in the values we have: (d/dx)[(f(x)/h(x))] at x=-2 = ((-8)*4 - 8*12) / (-8)^2Plug in values for quotient rule: Now we perform the calculations: (d/dx)[(f(x)/h(x))] at x=-2 = (-32 - 96) / 64Perform calculations: Simplify the numerator: (d/dx)[(f(x)/h(x))] at x=-2 = -128 / 64Simplify numerator: Finally, divide the numerator by the denominator: (d/dx)[(f(x)/h(x))] at x=-2 = -2

- Let $f$ be a function such that $f(-2)=8$ and $f^{\prime}(-2)=4$ . $\newline$ - Let $h$ be the function $h(x)=x^{3}$ . $\newline$ Evaluate $\frac{d}{d x}\left[\frac{f(x)}{h(x)}\right]$ at $x=-2$ .

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- Let f f f be a function such that f(−2)=8 f(-2)=8 f(−2)=8 and f′(−2)=4 f^{\prime}(-2)=4 f′(−2)=4.\newline- Let h h h be the function h(x)=x3 h(x)=x^{3} h(x)=x3.\newlineEvaluate ddx[f(x)h(x)] \frac{d}{d x}\left[\frac{f(x)}{h(x)}\right] dxd​[h(x)f(x)​] at x=−2 x=-2 x=−2.

- Let $f$ be a function such that $f(-2)=8$ and $f^{\prime}(-2)=4$ . $\newline$ - Let $h$ be the function $h(x)=x^{3}$ . $\newline$ Evaluate $\frac{d}{d x}\left[\frac{f(x)}{h(x)}\right]$ at $x=-2$ .