find nth derivative of $y = 3*e^{2x-1}$

asked by guest
on Jan 17, 2025 at 12:18 pm



MathBot Answer:

The first few derivatives of \(y = 3 {e}^{2 x - 1}\) with respect to \(x\) are: \[ \begin{aligned} \frac{d y}{d x} &= 6 e^{2 x - 1}\\ \frac{d^{2} y}{d x^{2}} &= 12 e^{2 x - 1}\\ \frac{d^{3} y}{d x^{3}} &= 24 e^{2 x - 1}\\ \frac{d^{4} y}{d x^{4}} &= 48 e^{2 x - 1} \end{aligned} \]


The formula for the \(n^\text{th}\) derivative is: \[ \frac{d^n}{d x^n} \left[ 3 e^{2 x - 1} \right] = 3 \cdot 2^{n} e^{2 x - 1} \]


In general, the \(n^{\text{nth}}\) derivative of \(y = 3 {e}^{2 x - 1}\) is given by the formula: \[ \frac{d^n}{d x^n} \left[ a e^{b x + c} \right] = a b^n e^{b x + c} \] In the case of \(3 e^{2 x - 1}\), \(a = 3\), \(b = 2\) and \(c = -1\).


\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).