Derivative + Integral Formulas

d/dx [k]

0

d/dx [f(x) ± g(x)]

f′(x) ± g′(x)

d/dx [k·f(x)]

k·f′(x)

d/dx [f(x)·g(x)]

g(x)·f′(x) + g’(x)·f(x)

d/dx [f(x)/g(x)]

[g(x)·f′(x) − f(x)·g′(x)] / [g(x)]² (domain: g(x) ≠ 0)

d/dx [f(g(x))]

f′(g(x)) · g′(x) (chain rule; domain: g(x) in domain of f)

d/dx [xⁿ]

n · xⁿ⁻¹

d/dx [uⁿ] (chain form)

n · uⁿ⁻¹ · u′ (domain: depends on ‘n’; for non-integer ‘n’ ensure ‘u’ stays in domain)

d/dx [sin(u)]

cos(u) · u′ (domain: all real u)

d/dx [cos(u)]

−sin(u) · u′ (domain: all real u)

d/dx [tan(u)]

sec²(u) · u′ (domain: u ≠ π/2 + kπ)

d/dx [cot(u)]

−csc²(u) · u′ (domain: u ≠ kπ)

d/dx [sec(u)]

sec(u)·tan(u) · u′ (domain: u ≠ π/2 + kπ)

d/dx [csc(u)]

−csc(u)·cot(u) · u′ (domain: u ≠ kπ)

d/dx [e^^{u}]

e^^{u} · u′ (domain: all real u)

d/dx [a^^{u}]

a^^{u}·ln(a) · u′ (domain: a>0)

d/dx [ln|u|]

u′ / u (domain: u ≠ 0)

d/dx [sin⁻¹(u)]

u′ / √(1 − u²) (domain: |u| < 1)

d/dx [cos⁻¹(u)]

− u′ / √(1 − u²) (domain: |u| < 1)

d/dx [tan⁻¹(u)]

u′ / (1 + u²) (domain: all real u)

d/dx [cot⁻¹(u)]

− u′ / (1 + u²) (domain: all real u)

d/dx [sec⁻¹(u)]

u′ / (|u|·√(u² − 1)) (domain: |u| > 1)

d/dx [csc⁻¹(u)]

− u′ / (|u|·√(u² − 1)) (domain: |u| > 1)

d/dx [log__a(u)]

u’/u·​​​ln(a) (domain: a > 0, a ≠ 1, u>0)

f’(c) (Use for specific ‘c’ value)

\lim_{x \to c} \frac{f(x) - f(c)}{x - c}

How do you check if a function is differentiable at x=c using the limit definition of the derivative?

1. Compute the left-hand derivative:
   f'_-(c) = \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c}

2. Compute the right-hand derivative:
   f'_+(c) = \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c}

3. Compare:

   • If f'_-(c) = f'_+(c), f is differentiable at c.

   • If they are not equal, f is not differentiable at c .

If f is differentiable at x = c, then…

f is continuous at x = c
However, the opposite is not always true.

Limit Definition of a Derivative

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

Where is a function not differentiable?

A function is not differentiable at points where it is not continuous, has a corner or cusp, a vertical tangent, or where the left-hand and right-hand derivatives are not equal.

Left-Sum

Σ(i=1 to n) f(a + (i-1)Δx) · Δx where Δx = (b-a)/n

Right-Sum

Σ(i=1 to n) f(a + iΔx) · Δx where Δx = (b-a)/n

Midpoint-Sum

Σ(i=1 to n) f(a + (i - 1/2)Δx) · Δx where Δx = (b-a)/n

Trapezoidal-Sum

Σ(i=1 to n) [(f(x_{i-1}) + f(x_i))/2] · Δx where Δx = (b-a)/n

∫(a to b) f(x)dx

F(b) - F(a)

Mean Value Theorem for Integrals

If f is continuous on [a,b], then there exists a number c in [a,b] such that… ∫(a to b) f(x)dx = f(c) · (b-a)

[Where f(c) is known as the average value of the function]

d/dx[∫(a to x) f(t)dt]

d/dx[∫(a to x) f(t)dt] = f(x)

d/dx[∫(a to u(x)) f(t)dt]

d/dx[∫(a to u(x)) f(t)dt] = f(u(x)) · u'(x)

Average Value

f(c) = (1/(b-a)) · ∫(a to b) f(x)dx

∫[1/((x-a)(x-b))]dx

∫[1/((x-a)(x-b)(x-c))]dx

∫[1/((x-a)(x-b))]dx = ∫[A/(x-a) + B/(x-b)]dx

∫[1/((x-a)(x-b)(x-c))]dx = ∫[A/(x-a) + B/(x-b) + C/(x-c)]dx

∫(1/x)dx

∫(1/u)du

∫(u'/u)dx

∫(1/x)dx = ln|x| + C

∫(1/u)du = ln|u| + C

∫(u'/u)dx = ln|u| + C

∫sin(u)du

-cos(u) + C

∫cos(u)du

sin(u) + C

∫tan(u)du

-ln|cos(u)| + C or ln|sec(u)| + C

∫cot(u)du

ln|sin(u)| + C

∫sec(u)du

ln|sec(u) + tan(u)| + C

∫csc(u)du

-ln|csc(u) + cot(u)| + C or ln|csc(u) - cot(u)| + C

∫(1/√(a² - u²))du

arcsin(u/a) + C

∫(1/√(a² - u²))du

-arccos(u/a) + C

∫(1/(a² + u²))du

(1/a)arctan(u/a) + C

∫(1/(a² + u²))du

-(1/a)arccot(u/a) + C

∫(1/(u√(u² - a²)))du

(1/a)arcsec(|u|/a) + C

∫(1/(u√(u² - a²)))du

-(1/a)arccsc(|u|/a) + C

Integration by Parts

∫udv=uv−∫vdu