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Common Derivatives



Constant Functions
Power Functions
Polynomials
Trigonometric Functions — Sine and Cosine
Trigonometric Functions — Tangent, Cotangent, Secant, Cosecant
Exponential Functions
Logarithmic Functions
Summary of Common Derivatives
Common Pitfalls
Common Derivatives FAQ



The Core Derivative Library


A small collection of derivative formulas covers the vast majority of differentiation work. Polynomials, trigonometric functions, exponentials, and logarithms appear constantly, and their derivatives should be available from memory. Every more complex derivative—a composition, a product, an implicitly defined relation—ultimately reduces to these building blocks combined through differentiation rules.

Each formula below is provable from the limit definition. For some, the proof is a direct algebraic manipulation of the difference quotient. For others, it requires special limits such as lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1. The proofs confirm the formulas; the formulas replace the proofs in daily computation.

Key Terms

Derivative— the formulas on this page give f′(x)f'(x) for standard functions
Higher-Order Derivative— repeated differentiation of these standard forms

See All Calculus Definitions →


Constant Functions


For any constant cc:

Constant Rule (Derivatives)
ddx[c]=0\frac{d}{dx}[c] = 0
Learn more about this formula: Constant Rule (Derivatives) →


Pitfall: Applying the power rule to a constant
Reading e3e^3 as a power and writing ddx[e3]=3e2\frac{d}{dx}[e^3] = 3e^2 treats a number as a function of xx. e3≈20.09e^3 \approx 20.09 is a constant, so its derivative is 00.

The graph of a constant function is a horizontal line. Every secant line through two points on it has slope zero, so the tangent line at every point also has slope zero. From the limit definition, the difference quotient c−ch=0\frac{c - c}{h} = 0 for all h≠0h \neq 0, and the limit is 00.

Under differentiation, constant terms vanish. In any sum f(x)+cf(x) + c, the constant contributes nothing to the derivative: ddx[f(x)+c]=f′(x)\frac{d}{dx}[f(x) + c] = f'(x).

Power Functions


For any real exponent nn:

Power Rule (Derivatives)
ddx[xn]=nxn−1\frac{d}{dx}[x^n] = nx^{n-1}
Learn more about this formula: Power Rule (Derivatives) →


The rule applies uniformly across exponent types. Positive integers: ddx[x5]=5x4\frac{d}{dx}[x^5] = 5x^4. Negative integers: ddx[x−3]=−3x−4\frac{d}{dx}[x^{-3}] = -3x^{-4}. Fractions: ddx[x1/2]=12x−1/2=12x\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}. Irrational exponents: ddx[xπ]=πxπ−1\frac{d}{dx}[x^\pi] = \pi x^{\pi - 1}.

For positive integer nn, the proof expands (x+h)n(x+h)^n using the binomial theorem. The leading term after cancellation is nxn−1hnx^{n-1}h, and dividing by hh and taking the limit gives nxn−1nx^{n-1}. All higher-order terms contain hh as a factor and vanish in the limit.

Pitfall: Lowering a negative exponent toward zero
Computing −3+1-3 + 1 instead of −3−1-3 - 1 turns ddx[x−3]\frac{d}{dx}[x^{-3}] into −3x−2-3x^{-2}, which is −0.75-0.75 at x=2x = 2. The rule always subtracts 11: the derivative is −3x−4-3x^{-4}, which is −0.1875-0.1875 at x=2x = 2.

Extension to negative and fractional exponents uses the quotient rule and the chain rule respectively, or alternatively logarithmic differentiation: writing xn=enln⁡xx^n = e^{n \ln x} and applying the chain rule gives ddx[xn]=enln⁡x⋅nx=xn⋅nx=nxn−1\frac{d}{dx}[x^n] = e^{n \ln x} \cdot \frac{n}{x} = x^n \cdot \frac{n}{x} = nx^{n-1}.
Exponent type Function Derivative
Positive integer x⁵ 5x⁴
Negative integer x⁻³ = 1/x³ −3x⁻⁴
Positive fraction x1/2 = √x (1/2)x−1/2 = 1/(2√x)
Negative fraction x−1/2 = 1/√x −(1/2)x−3/2
Irrational xπ π · xπ−1
-5 -4 -3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y f(x) = x² f′(x) = 2x
f(x) = x² and f′(x) = 2x, marked at x = 1

The upper curve is the function, the lower one its derivative, and at x = 1 the marker reads f′(1) = 2, the slope of the parabola there. The derivative is a straight line because bringing the exponent down leaves degree one: that is the power rule drawn out. Move the marker and watch the slope of the parabola track the height of the line on the derivative explorer.

Every polynomial derivative is this picture repeated term by term.

Polynomials


A polynomial p(x)=anxn+an−1xn−1+⋯+a1x+a0p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 is differentiated term by term using the power rule, constant multiple rule, and sum rule:

Polynomials
p′(x)=nanxn−1+(n−1)an−1xn−2+⋯+a1p'(x) = na_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1

Each term drops its degree by one. The constant term a0a_0 disappears. A polynomial of degree nn has a derivative of degree n−1n - 1.

Polynomials are differentiable at every real number. Their derivatives are again polynomials, so the process can be repeated indefinitely. After nn differentiations, a degree-nn polynomial becomes a constant n!⋅ann! \cdot a_n. One more differentiation yields zero, and all subsequent derivatives remain zero. This termination property distinguishes polynomials from transcendental functions, whose derivatives cycle or persist indefinitely.

Trigonometric Functions — Sine and Cosine


The two fundamental trigonometric derivatives are:

Derivative of Sine
ddx[sin⁡x]=cos⁡x\frac{d}{dx}[\sin x] = \cos x
Learn more about this formula: Derivative of Sine →

Derivative of Cosine
ddx[cos⁡x]=−sin⁡x\frac{d}{dx}[\cos x] = -\sin x
Learn more about this formula: Derivative of Cosine →


The sine derivative is proved from the limit definition. Expanding sin⁡(x+h)\sin(x + h) using the angle addition formula gives sin⁡xcos⁡h+cos⁡xsin⁡h\sin x \cos h + \cos x \sin h. The difference quotient becomes

Trigonometric Functions — Sine and Cosine
sin⁡x(cos⁡h−1)+cos⁡xsin⁡hh=sin⁡x⋅cos⁡h−1h+cos⁡x⋅sin⁡hh\frac{\sin x(\cos h - 1) + \cos x \sin h}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}

The special limits lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1 and lim⁡h→0cos⁡h−1h=0\lim_{h \to 0} \frac{\cos h - 1}{h} = 0 yield the result: cos⁡x\cos x.

Fact: Degrees would spoil the formula
Measured in degrees, ddx[sin⁡x∘]=π180cos⁡x∘≈0.01745cos⁡x∘\frac{d}{dx}[\sin x^\circ] = \frac{\pi}{180}\cos x^\circ \approx 0.01745\cos x^\circ, because sin⁡hh\frac{\sin h}{h} tends to 11 only when hh is in radians. Radians are the one unit that makes the extra factor exactly 11.

All the trigonometric formulas on this page measure xx in radians. On the unit circle, an angle of hh radians cuts off an arc of length exactly hh, and for small hh the height sin⁡h\sin h of the arc's endpoint is almost the arc itself. That is the geometric content of sin⁡hh→1\frac{\sin h}{h} \to 1.

The cosine derivative follows similarly, or by differentiating cos⁡x=sin⁡(π/2−x)\cos x = \sin(\pi/2 - x) using the chain rule. The negative sign in (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x is essential and a frequent source of error.

Repeated differentiation cycles with period four: sin⁡x→cos⁡x→−sin⁡x→−cos⁡x→sin⁡x\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x. This periodicity extends to higher-order derivatives: dndxn[sin⁡x]=sin⁡(x+nπ/2)\frac{d^n}{dx^n}[\sin x] = \sin(x + n\pi/2).
-5 -4 -3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y f(x) = sin(x) f′(x) = cos(x)
f(x) = sin x and f′(x) = cos x, marked at x = 1

The derivative curve peaks where the sine climbs fastest, at x = 0, and crosses zero where the sine tops out, at π/2: the cosine is the sine’s slope read off point by point. At x = 1 the marker reads cos 1 ≈ 0.54. Slide the marker to a peak of the sine and watch the derivative hit zero on the derivative explorer.

The quarter-period shift between a function and its derivative is the whole content of these two formulas.

Trigonometric Functions — Tangent, Cotangent, Secant, Cosecant


The remaining four trigonometric derivatives follow from sine and cosine via the quotient rule or rewriting in terms of sine and cosine.

Derivative of Tangent
ddx[tan⁡x]=sec⁡2x\frac{d}{dx}[\tan x] = \sec^2 x
Learn more about this formula: Derivative of Tangent →

Derivative of Cotangent
ddx[cot⁡x]=−csc⁡2x\frac{d}{dx}[\cot x] = -\csc^2 x
Learn more about this formula: Derivative of Cotangent →

Derivative of Secant
ddx[sec⁡x]=sec⁡xtan⁡x\frac{d}{dx}[\sec x] = \sec x \tan x
Learn more about this formula: Derivative of Secant →

Derivative of Cosecant
ddx[csc⁡x]=−csc⁡xcot⁡x\frac{d}{dx}[\csc x] = -\csc x \cot x
Learn more about this formula: Derivative of Cosecant →


For tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}, the quotient rule gives cos⁡x⋅cos⁡x−sin⁡x⋅(−sin⁡x)cos⁡2x=cos⁡2x+sin⁡2xcos⁡2x=1cos⁡2x=sec⁡2x\frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.

A pattern runs through the six derivatives: the cofunctions—cos⁡\cos, cot⁡\cot, csc⁡\csc—each carry a negative sign in their derivatives, while sin⁡\sin, tan⁡\tan, sec⁡\sec do not. This sign pattern is worth noting as a memory aid rather than a coincidence.

Each derivative is valid on the domain of the original function. The derivatives of tan⁡x\tan x and sec⁡x\sec x are undefined at x=π/2+kπx = \pi/2 + k\pi (where cos⁡x=0\cos x = 0). The derivatives of cot⁡x\cot x and csc⁡x\csc x are undefined at x=kπx = k\pi (where sin⁡x=0\sin x = 0).
Function Derivative Sign Where derivative is undefined
sin x cos x + nowhere (defined on all ℝ)
cos x −sin x − nowhere (defined on all ℝ)
tan x sec² x + x = π/2 + kπ
cot x −csc² x − x = kπ
sec x sec x · tan x + x = π/2 + kπ
csc x −csc x · cot x − x = kπ
Every trigonometric derivative comes from those of sine and cosine, through the quotient rule.

Exponential Functions


The natural exponential function has the unique property of being its own derivative:

Derivative of Natural Exponential
ddx[ex]=ex\frac{d}{dx}[e^x] = e^x
Learn more about this formula: Derivative of Natural Exponential →

THEOREM
No other function satisfies f′(x)=f(x)f'(x) = f(x) except constant multiples CexCe^x.

This self-replicating property is what makes ee the natural base for exponential functions.

The proof from the limit definition uses the special limit lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1. The difference quotient for exe^x factors as ex⋅eh−1he^x \cdot \frac{e^h - 1}{h}, and the limit gives ex⋅1=exe^x \cdot 1 = e^x.

For a general base a>0a > 0, a≠1a \neq 1:

Derivative of General Exponential
ddx[ax]=axln⁡a\frac{d}{dx}[a^x] = a^x \ln a
Learn more about this formula: Derivative of General Exponential →


Pitfall: Using the power rule on an exponential
Bringing the exponent down turns ddx[2x]\frac{d}{dx}[2^x] into x⋅2x−1x \cdot 2^{x-1}, which is 00 at x=0x = 0. The variable is in the exponent, so the derivative is 2xln⁡22^x \ln 2, which is ln⁡2≈0.693\ln 2 \approx 0.693 at x=0x = 0.

This follows from rewriting ax=exln⁡aa^x = e^{x \ln a} and applying the chain rule. The factor ln⁡a\ln a is constant—it scales the derivative. When a=ea = e, ln⁡a=1\ln a = 1 and the factor disappears, confirming the special role of base ee.
-5 -4 -3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y f(x) = eˣ f′(x) = eˣ
f(x) = eˣ and f′(x) = eˣ coincide

The two curves lie exactly on top of each other: at every x the slope of eˣ equals its height, so at x = 1 both readings are e ≈ 2.72. No other function has this property, which is why e is the natural base. Check any point and see the two readouts agree on the derivative explorer.

Every other exponential differs from its derivative only by the constant factor ln a.

Logarithmic Functions


The natural logarithm has derivative:

Derivative of Natural Logarithm
ddx[ln⁡x]=1x\frac{d}{dx}[\ln x] = \frac{1}{x}
Learn more about this formula: Derivative of Natural Logarithm →

This can be derived by implicit differentiation. If y=ln⁡xy = \ln x, then ey=xe^y = x. Differentiating both sides: eydydx=1e^y \frac{dy}{dx} = 1, so dydx=1ey=1x\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}.

Alternatively, the limit definition gives lim⁡h→0ln⁡(x+h)−ln⁡xh=lim⁡h→01hln⁡(x+hx)=lim⁡h→01hln⁡(1+hx)\lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h} = \lim_{h \to 0} \frac{1}{h}\ln\left(\frac{x+h}{x}\right) = \lim_{h \to 0} \frac{1}{h}\ln\left(1 + \frac{h}{x}\right), which evaluates to 1x\frac{1}{x} using the limit definition of ee.

For a general base a>0a > 0, a≠1a \neq 1:

Derivative of General Logarithm
ddx[log⁡ax]=1xln⁡a\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}
Learn more about this formula: Derivative of General Logarithm →


Pitfall: Dropping the factor ln a
Using 1x\frac{1}{x} as the derivative of log⁡10x\log_{10} x gives slope 11 at x=1x = 1. The factor 1ln⁡10\frac{1}{\ln 10} is missing: the derivative is 1xln⁡10\frac{1}{x \ln 10}, about 0.4340.434 at x=1x = 1.

This follows from the change of base formula log⁡ax=ln⁡xln⁡a\log_a x = \frac{\ln x}{\ln a} and the constant multiple rule. The natural logarithm gives the simplest derivative—another reason ee is the preferred base.

RULE
The function ln⁡∣x∣\ln|x| extends the domain to all x≠0x \neq 0, and its derivative is 1x\frac{1}{x} for both positive and negative xx.

This extended form appears frequently in integration.
-5 -4 -3 -2 -1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y f(x) = ln(x) f′(x) = 1/x
f(x) = ln x and f′(x) = 1/x, marked at x = 1

The logarithm climbs steeply near zero and flattens as x grows, and its derivative curve 1/x records exactly that: large on the left, small on the right, equal to 1 at x = 1 where the marker sits. Move the marker toward zero and watch the slope reading run away on the derivative explorer.

The reciprocal derivative is what makes the logarithm the antiderivative of 1/x.

Summary of Common Derivatives


The complete set of common derivative formulas:

Summary of Common Derivatives
ddx[c]=0ddx[xn]=nxn−1\frac{d}{dx}[c] = 0 \qquad \frac{d}{dx}[x^n] = nx^{n-1}

Summary of Common Derivatives
ddx[sin⁡x]=cos⁡xddx[cos⁡x]=−sin⁡x\frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x

Summary of Common Derivatives
ddx[tan⁡x]=sec⁡2xddx[cot⁡x]=−csc⁡2x\frac{d}{dx}[\tan x] = \sec^2 x \qquad \frac{d}{dx}[\cot x] = -\csc^2 x

Summary of Common Derivatives
ddx[sec⁡x]=sec⁡xtan⁡xddx[csc⁡x]=−csc⁡xcot⁡x\frac{d}{dx}[\sec x] = \sec x \tan x \qquad \frac{d}{dx}[\csc x] = -\csc x \cot x

Summary of Common Derivatives
ddx[ex]=exddx[ax]=axln⁡a\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[a^x] = a^x \ln a

Summary of Common Derivatives
ddx[ln⁡x]=1xddx[log⁡ax]=1xln⁡a\frac{d}{dx}[\ln x] = \frac{1}{x} \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}

These twelve formulas, combined with the differentiation rules, handle every explicit function built from powers, trigonometric functions, exponentials, and logarithms. Functions involving inverse trigonometric, hyperbolic, or piecewise definitions require the additional formulas collected in derivatives of special functions. The master reference card below collects everything in one place, grouped by function family for quick lookup.
Family Function f(x) Derivative f'(x)
Constant c 0
Power xn  (n ∈ ℝ) n · xn−1
Trigonometric sin x cos x
Trigonometric cos x −sin x
Trigonometric tan x sec² x
Trigonometric cot x −csc² x
Trigonometric sec x sec x · tan x
Trigonometric csc x −csc x · cot x
Exponential ex ex
Exponential ax  (a > 0, a ≠ 1) ax · ln a
Logarithmic ln x  (x > 0) 1 / x
Logarithmic loga x  (a > 0, a ≠ 1) 1 / (x · ln a)
DownloadPNG2×PDF
Before you read
Twelve derivatives in five families
Every derivative on this page belongs to one of five families. Learn the family and the members follow.
  • L0Common derivatives
    • L1Constant
      • L2c0
    • L1Power
      • L2xⁿn ∈ ℝn · xⁿ⁻¹
    • L1Trigonometric
      • L2sin xcos x
      • L2cos x−sin x
      • L2tan xsec² x
      • L2cot x−csc² x
      • L2sec xsec x · tan x
      • L2csc x−csc x · cot x
    • L1Exponential
      • L2eˣeˣ
      • L2aˣa > 0, a ≠ 1aˣ · ln a
    • L1Logarithmic
      • L2ln xx > 01 / x
      • L2logₐ xa > 0, a ≠ 11 / (x · ln a)
Takeaway
The building blocks
These twelve are the starting points; the derivative rules combine them into the derivative of almost any function.
Common derivatives by family·/calculus/derivatives/commonLearn Math Class

Common Pitfalls


Five mistakes account for most wrong answers when these formulas are applied. Four of them are marked where they happen in the sections above; the sign of the cosine derivative is the one the sine and cosine section warns against. Here each is worked through in full, with the wrong and the right working side by side.

Applying the power rule to a constant

A constant written as a power, such as e3e^3, π2\pi^2 or 545^4, is differentiated as if its base were the variable.

Wrong working: ddx[e3]=3e2\frac{d}{dx}[e^3] = 3e^2, by the power rule.

Right working: e3≈20.09e^3 \approx 20.09 is a fixed number; nothing in it depends on xx. By the constant rule, ddx[e3]=0\frac{d}{dx}[e^3] = 0.

Why it happens: the expression has the shape "something to a power", and the power rule is matched by shape. The rule is about xnx^n, where the base is the variable and the exponent is fixed.

How to check: look for xx. If the expression contains no xx, its graph is a horizontal line and the derivative is 00.

Lowering a negative exponent toward zero

With a negative exponent, 11 is added to it instead of subtracted, so the new exponent moves toward zero.

Wrong working: ddx[x−3]=−3x−3+1=−3x−2\frac{d}{dx}[x^{-3}] = -3x^{-3 + 1} = -3x^{-2}. At x=2x = 2 this gives −34=−0.75-\frac{3}{4} = -0.75.

Right working: the rule gives n−1n - 1, and −3−1=−4-3 - 1 = -4. So ddx[x−3]=−3x−4\frac{d}{dx}[x^{-3}] = -3x^{-4}, and at x=2x = 2 this is −316=−0.1875-\frac{3}{16} = -0.1875.

Why it happens: for positive exponents, subtracting 11 makes the exponent smaller in size, and that habit is carried over. For negative exponents, subtracting 11 makes the exponent larger in size.

How to check: compute a difference quotient. For f(x)=x−3f(x) = x^{-3}, f(2.001)−f(2)0.001≈−0.1873\frac{f(2.001) - f(2)}{0.001} \approx -0.1873, close to −0.1875-0.1875 and far from −0.75-0.75.

Dropping the minus sign on the cosine derivative

The derivative of cos⁡x\cos x is written as sin⁡x\sin x, without the minus sign, by analogy with ddx[sin⁡x]=cos⁡x\frac{d}{dx}[\sin x] = \cos x.

Wrong working: ddx[cos⁡x]=sin⁡x\frac{d}{dx}[\cos x] = \sin x, so the slope of cos⁡x\cos x at x=π2x = \frac{\pi}{2} is sin⁡π2=1\sin \frac{\pi}{2} = 1.

Right working: ddx[cos⁡x]=−sin⁡x\frac{d}{dx}[\cos x] = -\sin x, so the slope at x=π2x = \frac{\pi}{2} is −1-1.

Why it happens: the two formulas are learned as a pair, and the pair looks symmetric. It is not: going from sine to cosine keeps the sign, going from cosine to sine changes it.

How to check: think of the graph. Between 00 and π\pi the cosine falls from 11 to −1-1, so its slope there must be negative, and −sin⁡x-\sin x is negative on that interval.

Using the power rule on an exponential

In 2x2^x the variable is the exponent, but the power rule is applied as if it were xnx^n: the exponent is brought down and lowered by one.

Wrong working: ddx[2x]=x⋅2x−1\frac{d}{dx}[2^x] = x \cdot 2^{x - 1}. At x=0x = 0 this gives 00.

Right working: the exponential rule applies. ddx[2x]=2xln⁡2\frac{d}{dx}[2^x] = 2^x \ln 2, and at x=0x = 0 this is ln⁡2≈0.693\ln 2 \approx 0.693.

Why it happens: x2x^2 and 2x2^x use the same symbols, and the power rule is the first rule learned, so it is applied by habit.

How to check: compute a difference quotient at x=0x = 0. 20.001−200.001≈0.6934\frac{2^{0.001} - 2^0}{0.001} \approx 0.6934, which is not 00. An exponential with a base above 11 is always increasing, so its slope cannot be 00 anywhere.

Dropping the factor ln a

The derivative of log⁡ax\log_a x is written as 1x\frac{1}{x}, the formula for the natural logarithm, for every base.

Wrong working: ddx[log⁡10x]=1x\frac{d}{dx}[\log_{10} x] = \frac{1}{x}, so the slope at x=1x = 1 is 11.

Right working: log⁡10x=ln⁡xln⁡10\log_{10} x = \frac{\ln x}{\ln 10}, so ddx[log⁡10x]=1xln⁡10\frac{d}{dx}[\log_{10} x] = \frac{1}{x \ln 10}. At x=1x = 1 this is 1ln⁡10≈0.434\frac{1}{\ln 10} \approx 0.434.

Why it happens: 1x\frac{1}{x} is the logarithm derivative used most often, and the base is easy to overlook.

How to check: compute a difference quotient at x=1x = 1. log⁡101.001−log⁡1010.001≈0.4341\frac{\log_{10} 1.001 - \log_{10} 1}{0.001} \approx 0.4341, close to 0.4340.434 and not 11. Only for base ee, where ln⁡e=1\ln e = 1, is the derivative exactly 1x\frac{1}{x}.

Common Derivatives FAQ

What is the derivative of e^x?

+
It is e^x — the natural exponential is its own derivative. No other function has this property except constant multiples Ce^x, and that self-replication is what makes e the natural base. The proof factors the difference quotient as e^x · (e^h − 1)/h and applies the special limit (e^h − 1)/h → 1.Read more →

What is the derivative of ln x?

+
It is 1/x. The quickest derivation is implicit: if y = ln x then e^y = x, and differentiating gives e^y · dy/dx = 1, so dy/dx = 1/e^y = 1/x. The extended form ln|x| has the same derivative 1/x for every x ≠ 0, which is why it appears throughout integration.Read more →

What is the derivative of a^x?

+
For any base a > 0 with a ≠ 1, the derivative is a^x · ln a. It follows from rewriting a^x as e^(x ln a) and applying the chain rule, which leaves the constant factor ln a scaling the result. When a = e the factor becomes ln e = 1 and disappears — the reason base e is treated as the natural choice.Read more →

What is the derivative of tan x?

+
It is sec²x. Writing tan x as sin x / cos x and applying the quotient rule gives (cos x · cos x − sin x · (−sin x)) / cos²x, whose numerator collapses to cos²x + sin²x = 1 by the Pythagorean identity, leaving 1/cos²x = sec²x. The companion results are cot′ = −csc²x, sec′ = sec x tan x and csc′ = −csc x cot x.Read more →