Physics and Mathematics
Logarithmic Differentiation
1. Why Logarithmic Differentiation?
Some functions are difficult to differentiate directly, such as:
- Functions with variables in both base and exponent → [x^{x}], [(x^{2}+3)^{\sqrt{x}}]
- Products of many functions → [x(x+1)(x^{2}+1)]
- Quotients of complex expressions → [\dfrac{\sin x\cdot \sqrt{x}}{(x+1)^{3}}]
In such cases, taking logarithm on both sides simplifies the expression.
2. Key Steps
Let [y=f(x)].
- Take natural log on both sides:
[\ln y = \ln(f(x))] - Use log properties to simplify:
- Log of products → sum
$ \displaystyle \ln \left( {f\left( x \right).g\left( x \right)} \right)$ $=\ln \left( {f\left( x \right)} \right)+\ln \left( {g\left( x \right)} \right)$ - Log of quotients → difference
$ \displaystyle \ln \left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)$ $=\ln \left( {f\left( x \right)} \right)-\ln \left( {g\left( x \right)} \right)$ - Log of powers → exponent × log
$ \displaystyle \ln {{\left( {f\left( x \right)} \right)}^{{g\left( x \right)}}}$ $=g\left( x \right)\ln \left( {f\left( x \right)} \right)$
- Log of products → sum
- Now differentiate implicitly w.r.t [x]:
[\dfrac{d}{dx}(\ln y)][=\dfrac{1}{y}\dfrac{dy}{dx}] - Solve for [\dfrac{dy}{dx}]
3. Examples (Step-by-Step)
Example 1
Differentiate [y = x^{x}].
Step-by-Step Solution:
- Take ln on both sides:
[\ln y = \ln(x^{x})] - Use power rule of log:
[\ln y = x\ln x] - Differentiate implicitly:
[\dfrac{1}{y}\dfrac{dy}{dx}][ = \ln x + x\dfrac{1}{x}]
[\dfrac{1}{y}\dfrac{dy}{dx}][ = \ln x + 1] - Multiply both sides by [y=x^{x}]:
[\dfrac{dy}{dx}][ = x^{x}(\ln x + 1)]
Conclusion:
[\dfrac{dy}{dx} = x^{x}(\ln x + 1)]
Example 2
Differentiate [y = (x^{2}+3)^{5}].
Step-by-Step Solution:
- Take log:
[\ln y = \ln((x^{2}+3)^{5})] - Simplify:
[\ln y = 5\ln(x^{2}+3)] - Differentiate:
[\dfrac{1}{y}\dfrac{dy}{dx}][ = 5\cdot \dfrac{1}{x^{2}+3}\cdot 2x]
[= \dfrac{10x}{x^{2}+3}] - Multiply by [y=(x^{2}+3)^{5}]:
[\dfrac{dy}{dx}][ = (x^{2}+3)^{5}\cdot \dfrac{10x}{x^{2}+3}] - Simplify:
[\dfrac{dy}{dx} = 10x(x^{2}+3)^{4}]
Example 3
Differentiate [y][ = \dfrac{x\sqrt{x^{2}+1}}{(x+3)^{2}}].
Step-by-Step Solution:
- Apply ln on both sides:
[\ln y][ = \ln x + \ln(\sqrt{x^{2}+1}) – \ln((x+3)^{2})] - Simplify logs:
[\ln y][ = \ln x + \dfrac{1}{2}\ln(x^{2}+1) – 2\ln(x+3)] - Differentiate:
[\dfrac{1}{y}\dfrac{dy}{dx}][ = \dfrac{1}{x} + \dfrac{1}{2}\cdot \dfrac{2x}{x^{2}+1} – 2\cdot \dfrac{1}{x+3}]
[\dfrac{1}{y}\dfrac{dy}{dx}][ = \dfrac{1}{x} + \dfrac{x}{x^{2}+1} – \dfrac{2}{x+3}] - Multiply both sides by [y]:
[\dfrac{dy}{dx}][ = \dfrac{x\sqrt{x^{2}+1}}{(x+3)^{2}} \left( \dfrac{1}{x} + \dfrac{x}{x^{2}+1} – \dfrac{2}{x+3} \right)]
Final answer may remain unexpanded.
Example 4
Differentiate [y = (\sin x)^{\cos x}].
Step-by-Step Solution:
- Take ln:
[\ln y][ = \cos x \cdot \ln(\sin x)] - Differentiate:
[\dfrac{1}{y}\dfrac{dy}{dx}][ = -\sin x \cdot \ln(\sin x) + \cos x \cdot \dfrac{\cos x}{\sin x}]
But correction: apply product rule correctly ⬇
[\dfrac{1}{y}\dfrac{dy}{dx}][ = -\sin x \cdot \ln(\sin x) + \cos x \cdot \dfrac{\cos x}{\sin x}]
That is:
[\dfrac{1}{y}\dfrac{dy}{dx}][ = -\sin x \ln(\sin x) + \dfrac{\cos^{2}x}{\sin x}] - Multiply by [y=(\sin x)^{\cos x}]:
[\dfrac{dy}{dx}][ = (\sin x)^{\cos x}\left(-\sin x \ln(\sin x) + \dfrac{\cos^{2}x}{\sin x}\right)]
(We can simplify further, but this form is perfectly acceptable.)
Example 5
Differentiate [y = x^{\sin x}].
Step-by-Step Solution:
- Take log:
[\ln y][ = \sin x \cdot \ln x] - Differentiate using product rule:
[\dfrac{1}{y}\dfrac{dy}{dx}][ = \cos x \ln x + \sin x \cdot \dfrac{1}{x}] - Multiply both sides by [y=x^{\sin x}]:
[\dfrac{dy}{dx}][ = x^{\sin x}\left(\cos x \ln x + \dfrac{\sin x}{x}\right)]
4. Conceptual Questions with Solutions
1. Why do we take log on both sides in logarithmic differentiation?
Because logarithms convert **products into sums, quotients into differences and powers into multipliers**, making differentiation easier. Example: [y = x^{x}] becomes [\ln y = x\ln x], which is simpler to differentiate.
2. Why do we use implicit differentiation after taking log?
After logging, the variable [y] still appears inside [\ln y]. So we differentiate w.r.t [x] treating [y] as a function of [x]: [\dfrac{d}{dx}(\ln y) = \dfrac{1}{y}\dfrac{dy}{dx}].
3. When is logarithmic differentiation necessary?
When the function involves: – Variables as both base and exponent (e.g., [x^{x}]) – Complex products or quotients – Many factors or powers It simplifies the structure before differentiating.
4. Can logarithmic differentiation be used for normal functions like [x^{2}]?
Yes, it works. But it is not efficient since simpler rules exist. Use it only when required for simplification.
5. Why must the function inside log be positive?
Because the natural logarithm is **defined only for positive real values**. So if we apply it: [\ln y] implies [y > 0].
6. What happens if the given function is negative?
We rewrite using absolute value: [\ln|y|] This allows logarithmic differentiation even if [y] becomes negative in domain.
7. Why do we multiply both sides by [y] in the end?
Because differentiation of [\ln y] gives: [\dfrac{1}{y}\dfrac{dy}{dx}] So multiplying by [y] isolates [\dfrac{dy}{dx}].
8. Why is logarithmic differentiation useful for functions like [(x^{2}+1)^{x}]?
Because the exponent contains variable [x], so regular power rule doesn’t apply. Taking log converts the exponent into a product: [\ln y = x\ln(x^{2}+1)]
9. Does logarithmic differentiation always involve the quotient rule?
No. Even quotients become simpler after log: [\ln\Big(\dfrac{u}{v}\Big) = \ln u – \ln v] This avoids direct quotient rule.
10. Can we differentiate [y^{x}] if both are functions of [x]?
Yes. Logarithmic differentiation handles expressions like [y = u(x)^{v(x)}]. Take ln: [\ln y = v(x)\ln u(x)] and differentiate.
11. What rule is used after taking log of a product?
Product inside log becomes a sum: [\ln(uv) = \ln u + \ln v] Then differentiate each term separately.
12. Can logarithmic differentiation be combined with chain rule?
Yes. Many expressions require both: [\ln(\sin x)], [\ln(x^{2}+1)], etc. We apply chain rule while differentiating logs.
13. Is logarithmic differentiation helpful in series of products?
Very useful. Example: [y = u\cdot v\cdot w\cdot z\cdots] [\ln y = \ln u + \ln v + \ln w + \ln z + \cdots]
14. Why is simplification recommended before differentiating?
To reduce errors and complexity. Example: rewrite roots to exponents before taking logs.
15. How does logarithmic differentiation reduce long quotient rule expressions?
Because logs convert division into subtraction: [\ln\Big(\dfrac{u}{v}\Big)] [= \ln u – \ln v] Differentiating each term separately avoids messy quotient rule.
5. FAQ / Common Misconceptions
1. Misconception: Logarithmic differentiation is only for exponentials.
Reality: It helps for **products**, **quotients**, and **complicated powers**, not just exponentials.
2. Misconception: Taking log removes the variable from exponent.
No — log converts a power into multiplication but keeps the variable: [\ln(x^{x}) = x\ln x]
3. Misconception: We can take log on both sides without constraints.
Log requires the expression inside to be **positive**. Use [\ln|y|] if needed.
4. Misconception: After taking ln, function becomes simple forever.
No — extra steps are still needed like chain rule or simplification after differentiation.
5. Misconception: Logarithmic differentiation is slower.
For complicated expressions, it is actually **much faster** than product or quotient rules alone.
6. Misconception: We should never differentiate without simplifying the ln form.
Simplification is helpful but not mandatory. You can differentiate directly if comfortable.
7. Misconception: Chain rule is not part of logarithmic differentiation.
False. Differentiating log functions almost always invokes chain rule.
8. Misconception: Final answer can be left as [\dfrac{1}{y}\dfrac{dy}{dx}].
You must always multiply back by [y] to get [\dfrac{dy}{dx}].
9. Misconception: We can skip implicit differentiation step.
Incorrect — since [y] is a function of [x], implicit differentiation is **essential**.
10. Misconception: Log differentiation is optional even for [f(x)^{g(x)}] cases.
Not true — when both base and exponent depend on [x], it becomes **necessary**.
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