What is d/dx(e^(5x))?

Derivatives of Exponential Functions — eˣ, aˣ

§ 01 The Remarkable Derivative of e^x

Of all functions in mathematics, one stands alone: e^x is its own derivative. This single fact makes the natural exponential the engine of calculus, physics, finance, and biology.

The number e ≈ 2.71828… is defined precisely so that the function f(x) = e^x satisfies f′(x) = f(x). In other words, the rate at which e^x grows at any point equals its current value. No other base has this property: and that is exactly what makes e special.

Core Formula 1 — Natural Exponential

d/dx (e^x) = e^x

The derivative of e^x is itself. The slope at any point equals the height at that point.

Why Is This True? A Limit Argument

Using the definition of the derivative, we need to evaluate:

Derivative from First Principles

f′(x) = lim_h→0 [e^x+h − e^x] / h = e^x · lim_h→0 (e^h − 1) / h

The key limit: lim_h→0 (e^h − 1)/h = 1. This is effectively the definition of e.

The number e is defined as the unique base for which this limit equals exactly 1. Substituting back gives f′(x) = e^x · 1 = e^x.

Geometric Meaning At any point on the curve y = e^x, the tangent line has slope equal to the y-coordinate at that point. At x = 0, the slope is e⁰ = 1. At x = 2, the slope is e² ≈ 7.39. The function literally accelerates by its own height — the steeper it gets, the faster it gets steeper.

§ 02 Derivative of a^x — Any Base

What about exponentials with a base other than e, like 2^x or 10^x? We can rewrite any base using e:

Derivation

a^x = e^{x ln a}

Since ln(a) is a constant, differentiate using the chain rule:

d/dx (a^x) = e^{x ln a} · ln a = a^x · ln a

Core Formula 2 — General Exponential

d/dx (a^x) = a^x · ln a

Note: when a = e, ln(e) = 1, so this reduces to d/dx(e^x) = e^x ✓

Comparison of Common Bases

Function	Derivative	Multiplier ln(a)	Notes
e^x	e^x	ln(e) = 1	Unique: derivative = itself
2^x	2^x · ln 2	ln 2 ≈ 0.693	Grows slower than e^x
3^x	3^x · ln 3	ln 3 ≈ 1.099	Grows faster than e^x
10^x	10^x · ln 10	ln 10 ≈ 2.303	Common in engineering
(1/2)^x	(1/2)^x · ln(1/2)	ln(1/2) = −ln 2 ≈ −0.693	Negative multiplier → decay

Common Mistake: Confusing Power Rule with Exponential Rule The power rule differentiates xⁿ (variable base, constant exponent): d/dx(xⁿ) = nxⁿ⁻¹.
The exponential rule differentiates a^x (constant base, variable exponent): d/dx(a^x) = a^x·ln a.
These are completely different! d/dx(x³) = 3x² but d/dx(3^x) = 3^x·ln 3.

§ 03 Chain Rule with Exponentials — e^f(x) Forms

In practice you rarely see bare e^x. The exponent is almost always a function of x. The chain rule handles this: differentiate the outer exponential, then multiply by the derivative of the inner function.

Chain Rule Formula for e^f(x)

d/dx [e^f(x)] = e^f(x) · f′(x)

"The exponential survives unchanged, multiplied by the derivative of whatever is in the exponent."

Chain Rule Formula for a^f(x)

d/dx [a^f(x)] = a^f(x) · ln(a) · f′(x)

How to Apply the Chain Rule Step-by-Step

For d/dx[e^3x²]:

Identify the outer function: the exponential e^u where u = 3x².
Differentiate the outer: d/du(e^u) = e^u = e^3x². The exponential is unchanged.
Differentiate the inner: d/dx(3x²) = 6x.
Multiply: e^3x² · 6x = 6x e^3x².

d/dx[e^3x²] = 6x e^3x²

§ 04 12 Fully Worked Examples

Examples escalate from basic e^x through general bases, chain rule applications, product/quotient combinations, and implicit differentiation.

Example 01 Find d/dx(e^x)

This is the base case. Apply the fundamental formula directly.

The function is e^x. The exponent is simply x, so f(x) = x and f′(x) = 1.
Applying d/dx(e^x) = e^x · 1 = e^x.

d/dx(e^x) = e^x

Example 02 Find d/dx(e^5x)

The exponent is 5x: a linear function of x. Use the chain rule.

Outer function: e^u. Inner function: u = 5x.
d/du(e^u) = e^u = e^5x.
d/dx(5x) = 5.
Multiply: e^5x · 5.

d/dx(e^5x) = 5e^5x

Example 03 Find d/dx(e^x²+3x)

The exponent is a polynomial. Differentiate the exponent, multiply.

Outer: e^u → e^u. Inner: u = x² + 3x.
d/dx(x² + 3x) = 2x + 3.
Chain rule result: e^x²+3x · (2x + 3).

d/dx(e^x²+3x) = (2x + 3)e^x²+3x

Example 04 Find d/dx(3^x)

A non-e base. Use d/dx(a^x) = a^x·ln a with a = 3.

Identify a = 3. The formula gives 3^x · ln 3.
ln 3 ≈ 1.0986 is a positive constant, so the derivative is simply 3^x scaled by ln 3.

d/dx(3^x) = 3^x · ln 3

Example 05 Find d/dx(2^3x)

A non-e base with a compound exponent: combine the general formula with the chain rule.

Write as a^f(x) with a = 2, f(x) = 3x.
Formula: a^f(x)·ln(a)·f′(x).
f′(x) = d/dx(3x) = 3.
Substitute: 2^3x · ln 2 · 3 = 3 ln 2 · 2^3x.

d/dx(2^3x) = 3 ln 2 · 2^3x

Example 06 Find d/dx(e^{sin x})

The exponent is a trig function. The chain rule still applies.

Outer: e^u → e^u. Inner: u = sin x.
d/dx(sin x) = cos x.
Chain rule: e^{sin x} · cos x.

d/dx(e^{sin x}) = cos(x) · e^{sin x}

Example 07 Find d/dx(x² e^x)

A product of a polynomial and an exponential, use the product rule (uv)′ = u′v + uv′.

Let u = x² → u′ = 2x. Let v = e^x → v′ = e^x.
Product rule: u′v + uv′ = 2x · e^x + x² · e^x.
Factor out e^x: e^x(2x + x²) = x(x+2)e^x.

d/dx(x²e^x) = x(x + 2)e^x

Example 08 Find d/dx(e^x / x)

A quotient of exponential over polynomial, use the quotient rule (u/v)′ = (u′v − uv′)/v².

u = e^x → u′ = e^x. v = x → v′ = 1.
Quotient rule: (e^x · x − e^x · 1) / x².
Factor numerator: e^x(x − 1) / x².

d/dx(e^x/x) = e^x(x − 1) / x²

Example 09 Find d/dx(e^√x)

The exponent is a square root, equivalently x^1/2. Apply the chain rule.

Outer: e^u → e^u. Inner: u = x^1/2.
d/dx(x^1/2) = (1/2)x^−1/2 = 1/(2√x).
Chain rule: e^√x · 1/(2√x).

d/dx(e^√x) = e^√x / (2√x)

Example 10 Find d/dx(e^−x²)

This is the famous Gaussian bell-curve shape. The exponent is −x².

Outer: e^u → e^u. Inner: u = −x².
d/dx(−x²) = −2x.
Chain rule: e^−x² · (−2x).

d/dx(e^−x²) = −2x e^−x²

Note: the derivative is zero at x = 0 (the peak of the bell curve) and negative for x > 0 (decreasing), consistent with the shape of e^−x².

Example 11 Find d/dx[(e^x − e^−x) / 2] (hyperbolic sine)

This is sinh(x). Differentiate term by term.

d/dx(e^x/2) = e^x/2.
d/dx(−e^−x/2): inner function is −x, so d/dx(−x) = −1. Chain rule gives −(1/2)e^−x·(−1) = e^−x/2.
Add: e^x/2 + e^−x/2 = (e^x + e^−x)/2 = cosh(x).

d/dx[sinh x] = cosh x

Example 12 Find d/dx(e^xy) — implicit, differentiating w.r.t. x

Here xy is the exponent; x and y are both variables. Treat y as a function of x and use the chain rule plus product rule on the exponent.

Let u = xy. Then d/dx(u) requires the product rule: d/dx(xy) = y + x(dy/dx).
Chain rule: d/dx(e^xy) = e^xy · d/dx(xy).
Substitute: e^xy · [y + x(dy/dx)].

d/dx(e^xy) = e^xy(y + x dy/dx)

§ 05 Quick Reference — All Key Formulas

Function	Derivative	Condition / Note
e^x	e^x	The only self-derivative function
e^kx	k e^kx	k is any real constant
e^f(x)	f′(x) · e^f(x)	Chain rule; f must be differentiable
a^x	a^x · ln a	a > 0, a ≠ 1
a^f(x)	f′(x) · ln(a) · a^f(x)	Chain rule + general base
e^−x	−e^−x	Exponential decay; f(x) = −x, f′ = −1
x e^x	(x + 1)e^x	Product rule applied
e^x/xⁿ	e^x(x − n)/xⁿ⁺¹	Quotient rule, x ≠ 0

Memory Aid — The "e survives" principle Whenever you differentiate any expression of the form e^[something], the e^[something] is always present in the answer, multiplied by the derivative of [something]. The exponential can never be "destroyed" by differentiation — this is what makes it so useful in modelling growth and decay.

§ 06 Common Mistakes

Mistake 1 — Forgetting the chain rule Writing d/dx(e^3x) = e^3x instead of 3e^3x. Always ask: "Is the exponent just x, or something more complicated?" If it is more than x, you need the chain rule — multiply by the derivative of the exponent.

Mistake 2 — Applying the power rule to exponentials Writing d/dx(e^x) = xe^x−1. This confuses xⁿ (power rule) with e^x (exponential rule). In xⁿ the variable is the base. In e^x the variable is the exponent. These are completely different situations.

Mistake 3 — Wrong formula for a^x Writing d/dx(3^x) = x · 3^x−1 (power rule) or just 3^x (forgetting the ln a factor). The correct answer is 3^x · ln 3. Always include the ln(base) multiplier for non-e bases.

Mistake 4 — Dropping the chain rule factor in products When differentiating x · e^2x, students sometimes write x · 2e^2x and forget to add the second term from the product rule. The full answer using (uv)′ = u′v + uv′ is: 1 · e^2x + x · 2e^2x = e^2x(1 + 2x).

§ 07 10-Question Practice Quiz

Test your understanding. Enter your answer in the box, then press Check. Reveal the worked answer if you get stuck.

Score: 0 / 0

Question 1 · Fill in the blank

Find d/dx(e^7x). Enter the coefficient: __ · e^7x

Chain rule: multiply e^7x by the derivative of 7x.

The exponent is 7x. d/dx(7x) = 7.
Chain rule: e^7x · 7.

d/dx(e^7x) = 7e^7x

Question 2 · Multiple choice

What is d/dx(4^x)?

x · 4^x−1 4^x · ln 4 4^x 4 · ln x

Use d/dx(a^x) = a^x·ln a with a = 4.
Answer: 4^x·ln 4.

Question 3 · Fill in the blank

Find d/dx(e^−4x). Enter the coefficient: __ · e^−4x

Derivative of −4x is −4. The chain rule factor will be negative.

d/dx(−4x) = −4.
Chain rule: e^−4x · (−4).

d/dx(e^−4x) = −4e^−4x

Question 4 · Multiple choice

Find d/dx(e^x²).

e^x² 2x · e^x² x² · e^x²−1 2e^x²

Exponent = x². d/dx(x²) = 2x.
Chain rule: e^x² · 2x.

Answer: 2x e^x²

Question 5 · Fill in the blank

If f(x) = x · e^x, then f′(x) = e^x(x + __).

Product rule: u = x, v = e^x. Then u′ = 1, v′ = e^x.

Product rule: (x)′·e^x + x·(e^x)′ = e^x + xe^x.
Factor: e^x(1 + x).

f′(x) = e^x(x + 1)

Question 6 · Multiple choice

Which of the following is d/dx(e^{cos x})?

cos(x) · e^{cos x} e^{sin x} −sin(x) · e^{cos x} sin(x) · e^{cos x}

d/dx(cos x) = −sin x.
Chain rule: e^{cos x} · (−sin x).

Answer: −sin(x) e^{cos x}

Question 7 · Fill in the blank

Find d/dx(5^2x). Express as k · ln 5 · 5^2x. What is k?

d/dx(a^f(x)) = a^f(x)·ln(a)·f′(x). What is f′(x) when f(x) = 2x?

a = 5, f(x) = 2x, f′(x) = 2.
Formula: 5^2x · ln 5 · 2.

d/dx(5^2x) = 2 ln 5 · 5^2x, so k = 2

Question 8 · Multiple choice

What is d/dx(e^x/x²)?

e^x(x − 2)/x² e^x(x − 2)/x³ e^x(x + 2)/x³ e^x/x

u = e^x, u′ = e^x. v = x², v′ = 2x.
Quotient rule: (e^x·x² − e^x·2x) / x⁴ = e^x(x² − 2x)/x⁴.
Simplify: e^x·x(x − 2)/x⁴ = e^x(x − 2)/x³.

Question 9 · Fill in the blank

Find the derivative of e^−x²/2 (the standard normal PDF shape). Express as k·x·e^−x²/2. What is k?

d/dx(−x²/2) = −x. Chain rule gives e^−x²/2·(−x).

Exponent = −x²/2. d/dx(−x²/2) = −x.
Chain rule: e^−x²/2·(−x) = −x·e^−x²/2.

k = −1, so the answer is −x·e^−x²/2

Question 10 · Multiple choice

Which statement is true about d/dx(e^x)?

It equals xe^x−1 It equals e^x−1 It equals e^x, the same function It equals 1/e^x

d/dx(e^x) = e^x. The function is its own derivative.

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Quiz complete!

← Previous

Log Derivatives

Parametric Differentiation

§ 01 The Remarkable Derivative of ex

Why Is This True? A Limit Argument

§ 02 Derivative of ax — Any Base

Comparison of Common Bases

§ 03 Chain Rule with Exponentials — ef(x) Forms

How to Apply the Chain Rule Step-by-Step

§ 04 12 Fully Worked Examples

§ 05 Quick Reference — All Key Formulas

§ 06 Common Mistakes

§ 07 10-Question Practice Quiz

§ 01 The Remarkable Derivative of e^x

§ 02 Derivative of a^x — Any Base

§ 03 Chain Rule with Exponentials — e^f(x) Forms