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