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10. Demonstrate and explain how to find the derivative of the following functions. Be sure to explicitly denote which derivative rules (scalar multiple, sum/difference, quotient, etc.) you are using in your work. \( \begin{array}{l}\text { (h) } \\ h(y)=2 \cos (y)-e^{y}\end{array} \)

Ask by Grant Sanders. in the United States
Feb 26,2025

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Answer

The derivative of \( h(y) = 2 \cos(y) - e^y \) is \( h'(y) = -2 \sin(y) - e^y \).

Solution

To find the derivative of the function \( h(y) = 2 \cos(y) - e^y \), we will apply the following derivative rules: 1. **Scalar Multiple Rule**: If \( f(y) = c \cdot g(y) \), then \( f'(y) = c \cdot g'(y) \), where \( c \) is a constant. 2. **Sum/Difference Rule**: If \( f(y) = g(y) \pm k(y) \), then \( f'(y) = g'(y) \pm k'(y) \). 3. **Derivative of Cosine**: The derivative of \( \cos(y) \) is \( -\sin(y) \). 4. **Derivative of Exponential Function**: The derivative of \( e^y \) is \( e^y \). Now, let's differentiate \( h(y) \) step by step: 1. **Differentiate \( 2 \cos(y) \)**: - Using the Scalar Multiple Rule: The derivative of \( 2 \cos(y) \) is \( 2 \cdot (-\sin(y)) = -2 \sin(y) \). 2. **Differentiate \( -e^y \)**: - The derivative of \( -e^y \) is \( -1 \cdot e^y = -e^y \) (using the Scalar Multiple Rule). 3. **Combine the derivatives**: - Now, applying the Sum/Difference Rule, we combine the derivatives: \[ h'(y) = -2 \sin(y) - e^y \] Thus, the derivative of the function \( h(y) = 2 \cos(y) - e^y \) is: \[ h'(y) = -2 \sin(y) - e^y \] This result shows how we applied the derivative rules step by step to arrive at the final answer.

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To find the derivative of the function \( h(y) = 2 \cos(y) - e^y \), we can apply the sum/difference rule and the scalar multiple rule. According to the sum/difference rule, the derivative of a sum or difference of functions is the sum or difference of their derivatives. 1. First, differentiate \( 2 \cos(y) \): using the scalar multiple rule, we take the derivative of \( \cos(y) \), which is \(-\sin(y)\), and multiply by 2. Thus, \( \frac{d}{dy}(2 \cos(y)) = 2 \cdot (-\sin(y)) = -2\sin(y) \). 2. Second, differentiate \( -e^y \): since the derivative of \( e^y \) is itself, we have \( \frac{d}{dy}(-e^y) = -e^y \). Combining these results gives us \( h'(y) = -2\sin(y) - e^y \). So, the final answer is \( h'(y) = -2\sin(y) - e^y \).

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