Round #1079

Differentiate h(x) = cos(5x^3 - 2x) using the chain rule. Give the exact derivative.

To differentiate h(x) = cos(5x^3 - 2x) using the chain rule, we'll follow these steps: 1. Identify the outer function and the inner function. - Outer function: cos(u) - Inner function: u = 5x^3 - 2x 2. Differentiate the outer function with respect to u. - d(cos(u))/du = -sin(u) 3. Differentiate the inner function with respect to x. - d(5x^3 - 2x)/dx = d(5x^3)/dx - d(2x)/dx = 15x^2 - 2 4. Apply the chain rule by multiplying the derivative of the outer function by the derivative of the inner function. - h'(x) = d(cos(u))/du * d(u)/dx - h'(x) = -sin(u) * (15x^2 - 2) - Substitute back u = 5x^3 - 2x into the equation. - h'(x) = -sin(5x^3 - 2x) * (15x^2 - 2) So, the exact derivative of h(x) = cos(5x^3 - 2x) is: h'(x) = -(15x^2 - 2) * sin(5x^3 - 2x) Or, written in a more conventional format: h'(x) = (2 - 15x^2) * sin(5x^3 - 2x)

To differentiate `h(x) = cos(5x^3 - 2x)` using the chain rule, we first identify the outer function `f(u) = cos(u)` and the inner function `g(x) = 5x^3 - 2x`. The derivative of the outer function with respect to `u` is `f'(u) = -sin(u)`, and the derivative of the inner function with respect to `x` is `g'(x) = d(5x^3 - 2x)/dx = 15x^2 - 2`. Applying the chain rule, `(f∘g)'(x) = f'(g(x))·g'(x)`, we substitute `g(x)` into `f'(u)` and multiply by `g'(x)`: ```csharp using System; public class Program { public static void Main() { // Define the function h(x) = cos(5x^3 - 2x) Func<double, double> h = x => Math.Cos(5 * Math.Pow(x, 3) - 2 * x); // Define the derivative of h(x) using the chain rule Func<double, double> hDerivative = x => -Math.Sin(5 * Math.Pow(x, 3) - 2 * x) * (15 * Math.Pow(x, 2) - 2); // Print the derivative Console.WriteLine("h'(x) = -sin(5x^3 - 2x) * (15x^2 - 2)"); } } ``` The exact derivative of `h(x) = cos(5x^3 - 2x)` is `h'(x) = -sin(5x^3 - 2x) * (15x^2 - 2)`.