Exponential, trigonometric and logarithmic functions are not polynomials, yet each has a neat derivative formula of its own.
eˣ in particular comes back unchanged when you differentiate it, so at every point the slope of its tangent equals the height of the graph there.
That sin x differentiates to cos x, and cos x to −sin x, is simply a reading of how fast each graph rises and falls.
Move the point of tangency to compare slope and height on eˣ, and check that the slope of the tangent to sin x matches the height of cos x.
A Function That Is Its Own Derivative
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y = eˣ and its tangent: the red dashed line is the height, the white segment is a base of length 1
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👀 Height and slope are equal
①The gold curve is y = eˣ and the blue line is the tangent at the chosen point
②Since (eˣ)′ = eˣ, the slope of the tangent equals the height eˣ of the point
③So the tangent always meets the x-axis exactly 1 unit to the left of the point: it is the hypotenuse of a right triangle with base 1 and height eˣ
Derivatives of Trigonometric Functions
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The tangent to y = sin x and the height of y = cos x at the same x
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🔄 The slope of sin x and cos x
①When sin x is at its peak (x = π/2) the tangent is horizontal, and there cos x = 0
②Where sin x is increasing, cos x > 0; where it is decreasing, cos x < 0
③At x = 0, where sin x climbs most steeply, cos x takes its largest value, 1. A derivative is a rate of change, so the two match up this way
Formulas for Various Functions
Exponential functions
(eˣ)′ = eˣ, (aˣ)′ = aˣ ln a
With base e the derivative is the function itself; with base a, a factor ln a appears
Trigonometric functions
(sin x)′ = cos x, (cos x)′ = −sin x
sin x differentiates to cos x, and cos x to −sin x
Logarithmic functions
(ln x)' = 1x, (loga x)' = 1x ln a
The derivative of the natural log is 1/x; for base a, multiply by 1/ln a
The Other Trigonometric Functions
Derivatives of tan x and sec x
(tan x)′ = sec²x, (sec x)′ = sec x · tan x
Applying the quotient rule to tan x = sin x / cos x gives 1/cos²x = sec²x
Derivatives of cot x and csc x
(cot x)′ = −csc²x, (csc x)′ = −csc x · cot x
The derivatives of cot x and csc x carry a minus sign
💡 Ways to remember
①sin x differentiates to cos x with no sign change
②cos x differentiates to −sin x
③tan x differentiates to sec²x
④The functions whose names start with co, namely cos, cot and csc, have derivatives with a minus sign
Worked Examples
Example 1
Differentiate y = ex sin x.
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Use the product rule (fg)′ = f′g + fg′ with f = ex and g = sin x.
(ex sin x)′ = (ex)′ sin x + ex (sin x)′
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Substitute (ex)′ = ex and (sin x)′ = cos x.
ex sin x + ex cos x = ex(sin x + cos x)
▸ ex(sin x + cos x)
Since ex stays the same when differentiated, an expression with a factor ex only needs the product rule applied carefully.
Example 2
Differentiate y = ln(cos x).
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Differentiate the outer function ln u and multiply by the derivative of the inner u = cos x.
y' = 1cos x · (cos x)'
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Substitute (cos x)′ = −sin x.
y' = -sin xcos x = -tan x
▸ −tan x
Because (ln u)′ = u′/u, the derivative of the inner cos x, −sin x, ends up in the numerator. The chain rule is covered in detail in the next chapter.