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Use logarithmic diffentriation to find dy/dx: y=(3x^4-2)^5/(3x^3 4)^2

User Regisxp
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Using logarithmic differentiation to find dy/dx for y = (3x⁴ - 2)⁵/(3x - 4)², dy/dx = 6[10x³/(3x⁴ - 2) - 1/(3x - 4)](3x⁴ - 2)⁵/(3x - 4)²

To use logarithmic differentiation to find dy/dx: y = (3x⁴ - 2)⁵/(3x - 4)², we proceed as follows

Since y = (3x⁴ - 2)⁵/(3x - 4)² taking natural logarithm of both sides, we have that

㏑y = ㏑(3x⁴ - 2)⁵/(3x - 4)²

㏑y = ㏑(3x⁴ - 2)⁵ - ㏑(3x - 4)²

㏑y = 5㏑(3x⁴ - 2) - 2㏑(3x - 4)

Taking the derivative with respect to x of both sides, we have that

d㏑y/dx = d[5㏑(3x⁴ - 2) - 2㏑(3x - 4)]/dx

d㏑y/dy × dy/dx = d[5㏑(3x⁴ - 2)/dx - d[2㏑(3x - 4)]/dx

1/y × dy/dx = 5d㏑(3x⁴ - 2)/dx - 2d㏑(3x - 4)/dx

Let u = (3x⁴ - 2) and v = (3x - 4)

= 5d㏑u/dx - 2d㏑v/dx

= 5d㏑u/u × du/dx - 2d㏑v/v × dv/dx

= 5/u × du/dx - 2/v × dv/dx

du/dx = 12x³ and dv/dx = 3

So, substituting these into the equation, we have that

1/y × dy/dx = 5/u × 12x³ - 2/v × 3

1/y × dy/dx = 60x³/u - 6/v

1/y × dy/dx = 60x³/(3x⁴ - 2) - 6/(3x - 4)

dy/dx = [60x³/(3x⁴ - 2) - 6/(3x - 4)]y

dy/dx = 6[10x³/(3x⁴ - 2) - 1/(3x - 4)](3x⁴ - 2)⁵/(3x - 4)²

So, differentiating dy/dx = 6[10x³/(3x⁴ - 2) - 1/(3x - 4)](3x⁴ - 2)⁵/(3x - 4)²

User Theruss
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