Derivative Calculation: Finding Second Derivative at x = 0

This problem asks us to find the value of the second derivative of a given function \(y = e^{3x} + e^{-5x}\) at a specific point, \(x = 0\). To do this, we need to apply the rules of differentiation twice and then substitute the given value of \(x\).

First Derivative of the Function

The given function is \(y = e^{3x} + e^{-5x}\). We need to find the first derivative, \(\frac{dy}{dx}\).

Recall the rule for differentiating exponential functions: if \(f(x) = e^{ax}\), then \(\frac{df}{dx} = ae^{ax}\).

• For the first term, \(e^{3x}\), here \(a = 3\). So, the derivative is \(3e^{3x}\).
• For the second term, \(e^{-5x}\), here \(a = -5\). So, the derivative is \(-5e^{-5x}\).

Combining these, the first derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{d}{dx}(e^{3x}) + \frac{d}{dx}(e^{-5x}) \]

\[ \frac{dy}{dx} = 3e^{3x} + (-5)e^{-5x} \]

\[ \frac{dy}{dx} = 3e^{3x} - 5e^{-5x} \]

Second Derivative of the Function

Next, we need to find the second derivative, \(\frac{d^2y}{dx^2}\), by differentiating the first derivative \(\frac{dy}{dx}\) with respect to \(x\) again.

We have \(\frac{dy}{dx} = 3e^{3x} - 5e^{-5x}\). We will apply the same differentiation rule.

• For the first term, \(3e^{3x}\), the derivative will be \(3 \times (3e^{3x}) = 9e^{3x}\).
• For the second term, \(-5e^{-5x}\), the derivative will be \(-5 \times (-5e^{-5x}) = 25e^{-5x}\).

Combining these, the second derivative \(\frac{d^2y}{dx^2}\) is:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3e^{3x}) - \frac{d}{dx}(5e^{-5x}) \]

\[ \frac{d^2y}{dx^2} = 3(3e^{3x}) - 5(-5e^{-5x}) \]

\[ \frac{d^2y}{dx^2} = 9e^{3x} + 25e^{-5x} \]

Evaluating the Second Derivative at x = 0

Finally, we need to find the value of \(\frac{d^2y}{dx^2}\) when \(x = 0\). We substitute \(x = 0\) into the expression for the second derivative:

\[ \frac{d^2y}{dx^2}\Big|_{x=0} = 9e^{3(0)} + 25e^{-5(0)} \]

Simplify the exponents:

\[ \frac{d^2y}{dx^2}\Big|_{x=0} = 9e^{0} + 25e^{0} \]

Recall that any non-zero number raised to the power of 0 is 1, so \(e^0 = 1\).

\[ \frac{d^2y}{dx^2}\Big|_{x=0} = 9(1) + 25(1) \]

\[ \frac{d^2y}{dx^2}\Big|_{x=0} = 9 + 25 \]

\[ \frac{d^2y}{dx^2}\Big|_{x=0} = 34 \]

Thus, the value of \(\frac{d^2y}{dx^2}\) at \(x = 0\) is 34.