Given the function f(x)=cos(4sqrtx), find f^(')(x).

Identify Functions: We are given the function f(x) = cos(4sqrt(x)) and we need to find its derivative f'(x). 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 times the derivative of the inner function.Derivative of Outer Function: First, let's identify the outer function and the inner function. The outer function is cos(u), where u = 4sqrt(x), and the inner function is 4sqrt(x). We will need to take the derivative of both of these functions.Derivative of Inner Function: The derivative of the outer function cos(u) with respect to u is -sin(u). We will later substitute u with 4sqrt(x).Apply Chain Rule: The derivative of the inner function 4sqrt(x) with respect to x is 4 * (1/2) * x^(-1/2) = 2x^(-1/2) = 2/sqrt(x), using the power rule for derivatives.Simplify Final Answer: Now we apply the chain rule. The derivative of f(x) with respect to x is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This gives us f'(x) = -sin(4sqrt(x)) * (2/sqrt(x)).Simplify the expression for f'(x) to get the final answer. f'(x) = -2sin(4sqrt(x))/sqrt(x).

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Given the function $\newline$ $f(x)=\cos(4\sqrt{x}),$ find $\newline$ $f^{\prime}(x).$

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Algebra 1

Domain and range of square root functions: equations

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Q. Given the function

\newline

f(x)=\cos(4\sqrt{x}),

find

\newline

f^{\prime}(x).

Identify Functions: We are given the function $f(x) = \cos(4\sqrt{x})$ and we need to find its derivative $f'(x)$ . 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 times the derivative of the inner function.
Derivative of Outer Function: First, let's identify the outer function and the inner function. The outer function is $\cos(u)$ , where $u = 4\sqrt{x}$ , and the inner function is $4\sqrt{x}$ . We will need to take the derivative of both of these functions.
Derivative of Inner Function: The derivative of the outer function $\cos(u)$ with respect to $u$ is $-\sin(u)$ . We will later substitute $u$ with $4\sqrt{x}$ .
Apply Chain Rule: The derivative of the inner function $4\sqrt{x}$ with respect to $x$ is $4 \times (1/2) \times x^{(-1/2)} = 2x^{(-1/2)} = \frac{2}{\sqrt{x}}$ , using the power rule for derivatives.
Simplify Final Answer: Now we apply the chain rule. The derivative of $f(x)$ with respect to $x$ is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This gives us $f'(x) = -\sin(4\sqrt{x}) \times \left(\frac{2}{\sqrt{x}}\right)$ .
Simplify Final Answer: Now we apply the chain rule. The derivative of $f(x)$ with respect to $x$ is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This gives us $f'(x) = -\sin(4\sqrt{x}) \times \left(\frac{2}{\sqrt{x}}\right)$ . Simplify the expression for $f'(x)$ to get the final answer. $f'(x) = -\frac{2\sin(4\sqrt{x})}{\sqrt{x}}$ .