What is Logarithmic Differentiation?
Logarithmic differentiation is a technique where we take the natural log of both sides of y = f(x) before differentiating. It converts hard multiplications and powers into easy additions.
Power in variable
y = x^x or y = (sin x)^x
Long products
y = x² · sin x · eˣ · ln x
Complex fractions
y = (x²·sin x) / (eˣ·cos x)
Function raised to function
y = (f(x))^(g(x))
The Method — Step by Step
Take ln of both sides
Write: ln y = ln [f(x)]
Simplify using log properties
Use: ln(ab) = ln a + ln b, ln(a/b) = ln a − ln b, ln(aⁿ) = n ln a
Differentiate both sides w.r.t. x
LHS becomes: (1/y) · dy/dx (using chain rule)
Solve for dy/dx
Multiply both sides by y to get dy/dx = y · [...]
Substitute back y = f(x)
Replace y with the original expression to get the final answer.
⟹ dy/dx = y · d/dx [ln y]
Solved Examples
Take ln: ln y = ln(xˣ) = x · ln x
Differentiate: (1/y)·dy/dx = d/dx[x·ln x]
= 1·ln x + x·(1/x) = ln x + 1
dy/dx = y·(ln x + 1) = xˣ (ln x + 1)
Take ln: ln y = ln(x²) + ln(sin x) + ln(eˣ)
= 2 ln x + ln(sin x) + x
Differentiate: (1/y)·dy/dx = 2/x + cos x/sin x + 1
= 2/x + cot x + 1
dy/dx = y · (2/x + cot x + 1) = x²·sin x·eˣ · (2/x + cot x + 1)
Take ln: ln y = ½ln x + ½ln(sin x) − x
Differentiate: (1/y)·dy/dx = 1/(2x) + cos x/(2 sin x) − 1
= 1/(2x) + cot x/2 − 1
dy/dx = y · (1/(2x) + cot x/2 − 1)
Take ln: ln y = cos x · ln(sin x)
Differentiate using Product Rule on RHS:
(1/y)·dy/dx = −sin x·ln(sin x) + cos x·(cos x/sin x)
= −sin x·ln(sin x) + cos²x/sin x
dy/dx = (sin x)^(cos x) · [cos²x/sin x − sin x·ln(sin x)]
Key Takeaway
Logarithmic differentiation is the only practical method for functions like y = f(x)^g(x) where both base and power contain x.
The trick: d/dx[ln y] = (1/y)·dy/dx — always leads you to dy/dx.
💡 In exams, after applying log differentiation — always substitute y back in terms of x for the final answer. Don't leave dy/dx in terms of y.