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Q. then find
- Define Function: We need to find the derivative of the function with respect to , where is defined as . To do this, we will use the chain rule and the quotient rule for derivatives.
- Find : First, let's denote the inside function as , where . We will need to find the derivative of with respect to , which we will call .
- Apply Quotient Rule: To find , we will use the quotient rule: , where and . We need to find the derivatives of and with respect to .
- Apply Chain Rule: The derivative of with respect to is , using the chain rule. Therefore, the derivative of with respect to is , and the derivative of with respect to is also .
- Apply Chain Rule: Now we can apply the quotient rule:
- Apply Quotient Rule: Simplifying the numerator of , we get: .
- Find : After canceling terms in the numerator, we find that .
- Apply Chain Rule: Now we need to find the derivative of with respect to , which we will call , using the chain rule. Since , .
- Substitute u: Finally, we apply the chain rule to find : .
- Simplify Expression: Substitute back into the equation to get in terms of : \frac{dy}{dx} = \left(\frac{\(1\)}{\(2\)}\right)\left(\frac{\sec(\(2\)x) - \(1\)}{\sec(\(2\)x) + \(1\)}\right)^{-\frac{\(1\)}{\(2\)}} \times \left(-\frac{\(4\)\sec(\(2\)x)\tan(\(2\)x)}{(\sec(\(2\)x) + \(1\))^\(2\)}\right).
- Simplify Expression: Substitute \(u back into the equation to get in terms of : \frac{dy}{dx} = \frac{1}{2}\left(\frac{\sec(2x) - 1}{\sec(2x) + 1}\right)^{-\frac{1}{2}} \times \left(-\frac{4\sec(2x)\tan(2x)}{(\sec(2x) + 1)^2}\right)\. Simplify the expression to get the final answer: \$\frac{dy}{dx} = -2\sec(2x)\tan(2x) / \left(\frac{\sec(2x) - 1}{\sec(2x) + 1}\right)^{\frac{1}{2}} \times (\sec(2x) + 1)^2.
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