Verified answer

The derivative of the second term x³ is trivial:

d[x³]/dx = 3x²

One must use logarithmic differentiation on the first term:

let y = 3^(x² + 1/x) then ln(y) = (x² + 1/x)ln(3)

Differentiating both sides:

(1/y)dy/dx = (2x - 1/x²)ln(3)

Multiply both sides by y:

dy/dx = (2x - 1/x²)ln(3)y

Substitute 3^(x² + 1/x) for y:

dy/dx = (2x - 1/x²)ln(3)3^(x² + 1/x)

Add the second term:

dy/dx = (2x - 1/x²)ln(3)3^(x² + 1/x) + 3x²

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Note

d aᵁ

—— = aᵁ·ln(a)·u'

dx

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3^(x² + ¹ₓ̸)            ← This is aᵁ

So,

d[x³ + 3^(x² + ¹ₓ̸)]

———————– = 3x² + 3^(x² + ¹ₓ̸)•ln(3)•(2x - ¹ₓ̸²)            ← ANSWER

           dx

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