01 · Gradient at one
Find the tangent to y = xˣ at x = 1.
Hint
1¹ = 1 and ln 1 = 0.
Worked solution
The gradient is 1, and the point is (1, 1). The tangent is y − 1 = x − 1, or y = x.
Understand · explore · practise
Logarithmic differentiation
Differentiate x to the power x and other variable powers using logarithms. Separate base and exponent contributions, simplify products and quotients, and keep real-domain restrictions.
Before you startImplicit differentiation; chain rule; natural logarithm laws
01 / Let logarithms expose a changing exponent
The power rule alone does not handle a variable exponent.
If y = uᵛ with u > 0:
ln y = v ln u
y′/y = v′ ln u + v u′/u
The formula has two contributions: one from the changing exponent and one from the changing base. Multiply by y at the end to obtain y′. Here u and v are differentiable functions of x on the interval being considered.
Changing base, changing exponentExplore
u = x; v = x. Base term = 1; exponent term = ln x.
x = 1; y = 1; y′/y = 1; y′ = 1.
Both base and exponent vary. Add their relative contributions, then multiply by the function value.
All model inputs have a positive base. Negative bars mean a negative contribution. The bars show y′/y, while the gold tangent shows the actual derivative y′.
02 / Differentiate x to the power x
Both the base and the exponent depend on x.
y = xˣ, for x > 0Worked example
ln y = x ln x
Take natural logs on the positive domain.
y′/y = ln x + 1
Differentiate the right side with the product rule.
y′ = y(ln x + 1)
Multiply by y; the left derivative was y′/y, not y′.
y′ = xˣ(ln x + 1)
Replace y with the original function.
Watch the two contributions to the logarithmic derivative
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Distinguish constant and variable powers
Choose a rule by what is changing.
Compare three expressionsWorked example
y = x²: y′ = 2x
Only the base varies; the ordinary power rule applies.
y = 2ˣ: y′ = 2ˣ ln 2
Only the exponent varies; use the exponential derivative.
y = xˣ: y′ = xˣ(ln x + 1)
Both vary, so both contributions are needed.
02 · Find the missing term
A student writes d/dx(xˣ) = x·x^(x − 1) = xˣ. Why is that incomplete?
Hint
The power-rule calculation has treated the exponent as constant.
Worked solution
It captures only the changing-base contribution. The additional changing-exponent term is xˣ ln x, so the full derivative is xˣ(1 + ln x).
03 · A scaled exponent
Differentiate y = x^(2x), for x > 0.
Hint
ln y = 2x ln x.
Worked solution
y′/y = 2 ln x + 2, hence y′ = 2x^(2x)(ln x + 1).
04 / Differentiate a changing base and exponent
Use the product rule after taking logs.
y = (x + 2)ˣ, for x > −2Worked example
ln y = x ln(x + 2)
The positive-base condition gives the stated domain.
y′/y = ln(x + 2) + x/(x + 2)
Differentiate both factors.
y′ = (x + 2)ˣ[ln(x + 2) + x/(x + 2)]
Restore the original function.
04 · Gradient at zero
Find the derivative of (x + 2)ˣ at x = 0.
Hint
The function value is 2⁰ = 1.
Worked solution
The derivative is ln 2, since the base-change contribution x/(x + 2) vanishes at zero.
05 · Two nonlinear parts
Differentiate (3x + 1)^(x²) for x > −1/3.
Hint
Differentiate x² ln(3x + 1).
Worked solution
The derivative is (3x + 1)^(x²)[2x ln(3x + 1) + 3x²/(3x + 1)].
05 / Use trig functions in a variable power
Keep the positive-base domain and radian convention.
y = x^(sin x), for x > 0Worked example
ln y = sin x · ln x
Take natural logs.
y′/y = cos x ln x + sin x/x
The derivative of sin x assumes radians.
y′ = x^(sin x)[cos x ln x + sin x/x]
Multiply by the original y.
06 · Trig in the base
Differentiate (sin x)ˣ on 0 < x < π.
Hint
The sine is positive on this interval.
Worked solution
The derivative is (sin x)ˣ[ln(sin x) + x cot x]. The formula is justified on the stated open interval.
07 · An exponential simplification
Differentiate (eˣ)^(cos x) for real x.
Hint
Rewrite it as e^(x cos x).
Worked solution
The derivative is e^(x cos x)(cos x − x sin x). This agrees with logarithmic differentiation and the base is positive everywhere.
06 / Turn a product into a sum of logarithms
This can be shorter than repeated product and quotient rules.
y = x²(x + 1)³/(x + 2), for x > 0Worked example
ln y = 2 ln x + 3 ln(x + 1) − ln(x + 2)
All factors in these logarithms are positive on the chosen domain.
y′/y = 2/x + 3/(x + 1) − 1/(x + 2)
Differentiate each term.
y′ = [x²(x + 1)³/(x + 2)] [2/x + 3/(x + 1) − 1/(x + 2)]
A factored answer is often clearer than an expansion.
08 · Evaluate a product derivative
For this y, find y′ at x = 1.
Hint
First find y = 8/3 and the bracket 2 + 3/2 − 1/3.
Worked solution
The bracket is 19/6, so y′ = (8/3)(19/6) = 76/9.
09 · Roots and quotients
Differentiate y = √x (x + 2)²/(x + 1), for x > 0.
Hint
Use ln y = ½ ln x + 2 ln(x + 2) − ln(x + 1).
Worked solution
y′ = [√x (x + 2)²/(x + 1)] [1/(2x) + 2/(x + 2) − 1/(x + 1)].
07 / Use ln|y| on a nonzero branch
The absolute value allows a negative function value.
d/dx[ln|y|] = y′/y, where y ≠ 0
y = x²(x − 2), away from its zerosWorked example
ln|y| = 2 ln|x| + ln|x − 2|
Work on intervals excluding x = 0 and x = 2.
y′/y = 2/x + 1/(x − 2)
The log derivatives keep the signed denominators.
y′ = x²(x − 2)[2/x + 1/(x − 2)] = 3x² − 4x
This simplifies to the polynomial derivative away from the zeros.
At x = 0 and x = 2, differentiate the original polynomial
The derivative exists there: 0 and 4 respectively. The log step itself was not valid at those zeros.
10 · A negative value
For y = x²(x − 2), find the value and derivative at x = 1.
Hint
The value is negative, so ln y is not real. ln|y| is valid.
Worked solution
y = −1; y′ = −1. Logarithmic differentiation gives y′/y = 2 − 1 = 1, then y′ = (−1)·1 = −1.
11 · Do not divide by a zero
Why can the log formula for this polynomial not be substituted into at x = 2, although its derivative is 4 there?
Hint
The function value is zero.
Worked solution
ln|y| and y′/y are undefined at y = 0. Use the original polynomial or a justified limit of the simplified derivative; do not substitute into a formula containing 1/(x − 2).
08 / State where the logarithmic argument is valid
A variable real power is safely defined using a positive base.
For a general real exponent v(x), define u(x)^v(x) = exp[v(x) ln u(x)] on intervals where u(x) > 0. Negative bases may give isolated real values for certain rational exponents, but this does not create an open real domain for the same general logarithmic formula.
Do not claim that every power with a negative base is undefined: a fixed integer power such as x³ is defined for negative x. Choose the domain for the actual function and method.
12 · A shifted variable power
Give a positive-base interval and derivative for (x − 1)^(x + 1).
Hint
Require x − 1 > 0.
Worked solution
On x > 1, the derivative is (x − 1)^(x + 1)[ln(x − 1) + (x + 1)/(x − 1)]. No claim is made about an extension at or below x = 1.
09 / Use the sign of the logarithmic derivative
For positive y, y′ and y′/y have the same sign.
Find the minimum of y = xˣ on x > 0Worked example
y′ = xˣ(ln x + 1)
The factor xˣ is positive.
ln x + 1 = 0 gives x = e⁻¹
There is exactly one stationary input.
y′ < 0 before e⁻¹ and y′ > 0 after e⁻¹
The function decreases then increases.
Minimum point: (e⁻¹, e^(−1/e))
The minimum is attained on the positive domain.
13 · A related minimum
Find the stationary point of y = x^(2x), x > 0, and classify it.
Hint
The sign comes from 2(ln x + 1).
Worked solution
The minimum is at (e⁻¹, e^(−2/e)). The derivative changes from negative to positive.
10 / Separate a relative rate from an absolute rate
The logarithmic derivative still needs the function value.
At x = 2 for y = xˣWorked example
y = 4
Evaluate the original function.
y′/y = ln 2 + 1
This is the relative rate per unit x.
y′ = 4(ln 2 + 1)
Multiply to obtain the actual gradient.
14 · Spot the unfinished answer
For y = (x + 1)ˣ, a solution stops at ln(x + 1) + x/(x + 1). What quantity has it found?
Hint
It came from differentiating ln y.
Worked solution
It has found y′/y. Multiply by (x + 1)ˣ to obtain y′, on x > −1.
15 · Relative and absolute rates
If y = 6 and y′/y = −0.4 at a point, find y′.
Hint
Multiply the relative rate by y.
Worked solution
y′ = 6(−0.4) = −2.4. The two quantities have different meanings.
11 / Take logs, differentiate and restore y
Check the domain before applying a log law.
- For a variable power, separate the contributions of base and exponent.
- Use product rules after taking logs when both factors vary.
- The derivative of ln y is y′/y.
- Multiply by the original y to finish.
- Use ln|y| on nonzero branches when signs matter.
- Check zeros and excluded points directly from the original function.
16 · A final variable power
Differentiate y = (x² + 1)^(2x) for real x.
Hint
The base is positive for every real x.
Worked solution
y′ = (x² + 1)^(2x)[2 ln(x² + 1) + 4x²/(x² + 1)].
Section 1 of 11 · Let logarithms expose a changing exponent