Differentiation: chain, product and quotient rules flashcards

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Composite function (function of a function)
A function built by applying one function to the output of another, e.g. (4x − 3)⁵ or e^(sin x): an outside function applied to an inside function.
Chain rule
dy/dx = (dy/du) × (du/dx): differentiate the outside, keep the inside, multiply by the derivative of the inside.
Stationary point of inflexion
A point where dy/dx = 0 but dy/dx has the same sign on both sides, so the curve is neither a maximum nor a minimum there.
Vertical tangent
A tangent parallel to the y-axis; dy/dx is undefined there, which happens where dx/dy = 0.
Radian measure (why it matters here)
The derivatives of sin, cos, tan, sec, cosec and cot are only true with the angle in radians.
d/dx (e^(kx))
k e^(kx). (No — learn it)
d/dx (ln kx), kx > 0
1/x (the k disappears, because ln kx = ln k + ln x). (No — learn it)
d/dx (sin kx) and d/dx (cos kx)
k cos kx and −k sin kx. (No — learn it)
d/dx (tan kx)
k sec² kx. (Yes)
d/dx (sec x), d/dx (cosec x), d/dx (cot x)
sec x tan x; −cosec x cot x; −cosec² x. (Yes)
Product rule
d/dx (uv) = u′v + uv′. (No — learn it)
Quotient rule
d/dx (u/v) = (u′v − uv′)/v². (Yes)
d/dx [ln f(x)]
f′(x)/f(x). (No — derive it)
dy/dx when x is given as a function of y
dy/dx = 1 ÷ (dx/dy), wherever dx/dy ≠ 0. (No — learn it)
How are the graphs of y = ln x and y = ln 3x related, and what does that say about their gradients?
y = ln 3x is y = ln x translated up by ln 3, so at every x both have the same gradient, 1/x.
Shape of y = (ln x)/x for x > 0
Crosses the x-axis at (1, 0), rises to a maximum at (e, 1/e), then decreases towards the x-axis.
Gradient of x = tan 2y at the origin and far from it
dy/dx = 1/(2(1 + x²)): ½ at the origin, tending to 0 as the curve flattens towards y = ±π/4.
Graph of y = f′(x) at a stationary point of inflexion of y = f(x)
It touches the x-axis there without crossing it (f′ has the same sign on both sides).
Why draw tangents and normals with equal scales?
With equal scales on both axes, perpendicular lines look perpendicular; otherwise the right angle is distorted.
Why does e^x appear as a factor in the derivative of x² e^x, and why does that help?
Both product-rule terms contain e^x. Since e^x > 0, setting dy/dx = 0 reduces to the polynomial factor.
How do you prove d/dx (sec x) = sec x tan x?
Write sec x = (cos x)⁻¹; chain rule gives −(cos x)⁻² × (−sin x) = sin x/cos² x = sec x tan x.
How does the quotient rule give d/dx (tan x)?
tan x = sin x/cos x, so (cos x · cos x − sin x · (−sin x))/cos² x = 1/cos² x = sec² x.
What do you do if d²y/dx² = 0 at a stationary point?
The test fails. Check the sign of dy/dx just either side: + to − maximum, − to + minimum, same sign: point of inflexion.
After using dy/dx = 1/(dx/dy), how do you get the answer in terms of x?
Replace the y-expression using the original equation or an identity, e.g. sec² 2y = 1 + tan² 2y = 1 + x² for x = tan 2y.
Differentiate (3x² + 1)⁴.
4(3x² + 1)³ × 6x = 24x(3x² + 1)³.
Differentiate ln(x² + 5).
2x/(x² + 5).
Differentiate x sin 2x.
Product rule: sin 2x + 2x cos 2x.
Differentiate e^(3x)/x.
Quotient rule: (3x e^(3x) − e^(3x))/x² = e^(3x)(3x − 1)/x².
Differentiate e^(sin x).
cos x · e^(sin x).
Differentiate cosec 3x.
−3 cosec 3x cot 3x.
Stationary point of y = x e^(−2x)?
dy/dx = (1 − 2x)e^(−2x) = 0 ⇒ x = ½; point (½, 1/(2e)).
x = cos² y. Find dy/dx in terms of y.
dx/dy = −2 sin y cos y = −sin 2y, so dy/dx = −1/sin 2y = −cosec 2y.
Tangent to y = e^x sin x at the origin?
dy/dx = e^x(sin x + cos x) = 1 at x = 0, so the tangent is y = x.
d/dx (ln 6x) — the common slip
It is 1/x, not 6/x. Only 6 ln x differentiates to 6/x.
Order in the quotient rule numerator
Bottom × derivative of top, minus top × derivative of bottom: (vu′ − uv′). Swapping them flips the sign.
Cancelling e^(…) or cos x when solving dy/dx = 0
State that e^(…) > 0 (never zero); for a trig factor, solve factor = 0 separately rather than dividing it away.
Calculator mode for gradients of trig curves
Radians. In degrees every derivative would need an extra factor π/180.
“Show that d/dx (cot x) = −cosec² x”
Quoting the booklet is not a proof: write cot x = cos x/sin x, use the quotient rule and simplify with sin² x + cos² x = 1.

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