When simplifying logarithmic expressions, always use properties such as \( \log_e(a^b) = b \log_e(a) \) to simplify the equation. Also, remember that \( \log_e(e^z) = z \), which simplifies logarithmic expressions with exponents. In cases where you have an exponential function like \( t = e^{2x} \), this property makes differentiation much easier. Finally, remember that the derivative of a linear function like \( 4x \) is a constant, and its second derivative will be zero.
First, simplify \( y = \log_e(t^2) \) as follows:
\( y = 2 \log_e(t) \)
Since \( t = e^{2x} \), we have:
\( \log_e(t) = 2x \implies y = 2 \cdot 2x = 4x \)
Now, taking the first derivative with respect to \( x \):
\( \frac{dy}{dx} = 4 \)
Then, taking the second derivative with respect to \( x \):
\( \frac{d^2y}{dx^2} = 0 \)
Thus, the value of \( \frac{d^2y}{dx^2} \) is 0.
First, simplify \( y = \log_e(t^2) \) as follows:
Using the logarithmic property \( \log_e(a^b) = b \log_e(a) \), we can simplify \( \log_e(t^2) \) as:
\[ y = 2 \log_e(t) \]
Step 1: Substitute the value of \( t \):
We are given that \( t = e^{2x} \), so we substitute \( t \) into the expression:
\[ y = 2 \log_e(e^{2x}) \]
Step 2: Apply the property of logarithms:
We know that \( \log_e(e^z) = z \), so applying this property gives:
\[ y = 2 \cdot 2x = 4x \]
Step 3: Find the first derivative of \( y \) with respect to \( x \):
Now that we have \( y = 4x \), we can differentiate it with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(4x) = 4 \]
Step 4: Find the second derivative of \( y \) with respect to \( x \):
The first derivative is a constant, so the second derivative is:
\[ \frac{d^2y}{dx^2} = 0 \]
Conclusion: Thus, the value of \( \frac{d^2y}{dx^2} \) is 0.