If `y=ae^(x)+be^(-x)+c," then "y'''=`

Options

A. y
B. 0
C. y'
D. y''

Answer

To find the third derivative of the function \( y = ae^x + be^{-x} + c \), we will go through the differentiation step by step.

### Step 1: Find the first derivative \( y' \)
Given:
\[ y = ae^x + be^{-x} + c \]
To find the first derivative \( y' \), we differentiate \( y \) with respect to \( x \):
\[ y' = \frac{d}{dx}(ae^x) + \frac{d}{dx}(be^{-x}) + \frac{d}{dx}(c) \]
Using the derivatives of \( e^x \) and \( e^{-x} \):
\[ y' = a e^x + b \left(-e^{-x}\right) + 0 \]
Thus, we have:
\[ y' = a e^x - b e^{-x} \]

### Step 2: Find the second derivative \( y'' \)
Now we differentiate \( y' \) to find the second derivative \( y'' \):
\[ y'' = \frac{d}{dx}(a e^x - b e^{-x}) \]
Differentiating each term:
\[ y'' = a e^x + b e^{-x} \]

### Step 3: Find the third derivative \( y''' \)
Now we differentiate \( y'' \) to find the third derivative \( y''' \):
\[ y''' = \frac{d}{dx}(a e^x + b e^{-x}) \]
Differentiating each term:
\[ y''' = a e^x - b e^{-x} \]

### Conclusion
Notice that \( y''' = y' \). Therefore, we conclude that:
\[ y''' = a e^x - b e^{-x} \]

### Final Answer
\[ y''' = y' \]

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