Derivative of ln(y)- Calculus Differentiation
What Is the Derivative of ln(y)?
The derivative of ln(y) with respect to x is (1/y) · dy/dx. That's it. There's no magic here.
You're applying the chain rule. The natural log function differentiates to 1/y, and then you multiply by the derivative of whatever's inside—which is y itself.
If you're differentiating ln(y) with respect to y directly, the answer is simply 1/y. But most calculus problems involve implicit differentiation, which means you're working with multiple variables.
Why the Chain Rule Matters Here
ln(y) isn't a simple function. The input to the logarithm is y, which itself changes with x. You can't just slap down 1/y and call it done.
The chain rule says:
d/dx[f(g(x))] = f'(g(x)) · g'(x)
For ln(y), treat the outer function as ln(u) where u = y. Then:
- Outer function: ln(u) → derivative is 1/u
- Inner function: y → derivative is dy/dx
- Multiply them: (1/y) · dy/dx
This is what most textbooks call implicit differentiation. You're not finding how ln(y) changes with y—you're finding how it changes with x when y itself is a function of x.
The Formula You Actually Need
Here's the clean version you should memorize:
| Function | Derivative | Condition |
|---|---|---|
| d/dx[ln(y)] | (1/y) · dy/dx | y is a function of x |
| d/dy[ln(y)] | 1/y | Taking derivative with respect to y |
| d/dx[ln(f(x))] | f'(x)/f(x) | f(x) is the inner function |
The third row is the most useful in practice. When you have ln(something), the derivative becomes (derivative of something) divided by (something itself).
Step-by-Step: How to Actually Do This
Example 1: Basic Implicit Differentiation
Find d/dx[ln(y)] when y = x³ + 2x
Step 1: Apply the chain rule formula
d/dx[ln(y)] = (1/y) · dy/dx
Step 2: Find dy/dx
dy/dx = 3x² + 2
Step 3: Substitute
d/dx[ln(y)] = (1/(x³ + 2x)) · (3x² + 2)
Simplify if the problem asks for it.
Example 2: Solving for dy/dx
Given: ln(y) = x² + 3y
Find dy/dx.
Step 1: Differentiate both sides
(1/y) · dy/dx = 2x + 3 · dy/dx
Step 2: Collect dy/dx terms on one side
(1/y) · dy/dx - 3 · dy/dx = 2x
Step 3: Factor out dy/dx
dy/dx · (1/y - 3) = 2x
Step 4: Solve
dy/dx = 2x / (1/y - 3) = 2xy / (1 - 3y)
That's your answer. No motivational quotes about persistence.
Common Mistakes That Will Cost You Points
- Forgetting the chain rule entirely. Some students write d/dx[ln(y)] = 1/y. Wrong. That's only correct when differentiating with respect to y.
- Dropping the dy/dx term. If y is a function of x, you must track how y changes with x.
- Not simplifying. Leaving answers as (1/y)·dy/dx when you've already defined dy/dx is lazy. Substitute or solve.
- Domain errors. ln(y) is only defined for y > 0. If your answer gives y ≤ 0, something went wrong.
Quick Reference Table
| Function | Derivative |
|---|---|
| ln(x) | 1/x |
| ln(f(x)) | f'(x)/f(x) |
| ln(y) where y = g(x) | g'(x)/g(x) |
| e^ln(y) | y (simplifies completely) |
| ln(e^x) | 1 (simplifies to x, derivative is 1) |
When You'll Actually Use This
Logarithmic differentiation shows up in:
- Finding derivatives of products with many factors like x⁵(x+1)⁴/(x-2)³
- Calculus problems involving exponential growth and decay
- Economics problems with elasticities
- Any situation where the variable appears both as a base and exponent
For products and quotients with mixed exponents, take ln of both sides first, differentiate, then solve. It turns complicated quotient rules into simple addition and subtraction.
Bottom Line
The derivative of ln(y) with respect to x is (1/y) · dy/dx. Apply the chain rule. Track your inner derivative. Don't forget that y is a function of x unless told otherwise.
That's the whole thing. No more to memorize.