CUET UG 2025 — Mathematics Calculus

Given $y = 5e^{2x} + 4e^{3x}$y=5e2x+4e3x

When differentiating $e^{ax}$eax, the derivative is $a \cdot e^{ax}$a⋅eax

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Finding the first derivative:

For $5e^{2x}$5e2x: the derivative is $5 \times 2 \times e^{2x} = 10e^{2x}$5×2×e2x=10e2x

For $4e^{3x}$4e3x: the derivative is $4 \times 3 \times e^{3x} = 12e^{3x}$4×3×e3x=12e3x

$$\frac{dy}{dx} = 10e^{2x} + 12e^{3x}$$dxdy​=10e2x+12e3x

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Finding the second derivative:

For $10e^{2x}$10e2x: the derivative is $10 \times 2 \times e^{2x} = 20e^{2x}$10×2×e2x=20e2x

For $12e^{3x}$12e3x: the derivative is $12 \times 3 \times e^{3x} = 36e^{3x}$12×3×e3x=36e3x

$$\frac{d^2y}{dx^2} = 20e^{2x} + 36e^{3x}$$dx2d2y​=20e2x+36e3x

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Factoring the result:

$$\frac{d^2y}{dx^2} = 20e^{2x} + 36e^{3x}$$dx2d2y​=20e2x+36e3x

$$\frac{d^2y}{dx^2} = 4(5e^{2x} + 9e^{3x})$$dx2d2y​=4(5e2x+9e3x)

Therefore, $\frac{d^2y}{dx^2} = 4(5e^{2x} + 9e^{3x})$dx2d2y​=4(5e2x+9e3x)