However tangled a function looks, there are only a few routes to its derivative.
When one function sits inside another, the chain rule differentiates the outside and multiplies by the derivative of the inside; when y is hard to solve for, implicit differentiation differentiates both sides of the equation as written; and when a point's position depends on a parameter t, dividing dy/dt by dx/dt gives the slope.
In the first picture, move x and the slope of the red tangent to sin 2x always matches the height of the blue curve 2cos 2x at the same x.
In the second, a rolling wheel traces a cycloid, and you can check that the tangent always passes through the top of the wheel.
The Chain Rule for Composite Functions
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y = sin 2x and its derivative y′ = 2cos 2x: the red slope equals the blue height
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🔗 Where the factor 2 comes from
①sin 2x is a composite: the outer function sin applied to the inner function 2x
②When x grows a little, 2x grows twice as much, and the value of sin changes by cos 2x times that increase. So the derivative is cos 2x multiplied by 2, which is 2cos 2x
③The gold curve is sin x squeezed to half its width, so it climbs and falls twice as fast. That factor 2 is why the blue curve reaches ±2
Chain rule
{f(g(x))}' = f'(g(x)) · g'(x)
The derivative of the outer function times the derivative of the inner function
Implicit Differentiation
Implicit differentiation of the circle x² + y² = r²
x² + y² = r² → 2x + 2y·y' = 0 → y' = −xy
Treat y as a function of x and differentiate both sides with respect to x. By the chain rule, y² becomes 2y · y′
🔄 Steps of implicit differentiation
①You do not have to solve for y = f(x) first. Even a circle, where one x gives two values of y, has a slope at every point
②Each time you differentiate a term containing y, the chain rule attaches a factor y′
③After differentiating both sides, solve for y′. For the circle y′ = −x/y; multiplied by the radius slope y/x it gives −1, which matches the fact that the tangent is perpendicular to the radius
Differentiating Parametric Curves
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A cycloid traced by a rolling wheel of radius 1, with its tangent
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Parametric differentiation
x = f(t), y = g(t) → dydx = dy/dtdx/dt = g'(t)f'(t)
Differentiate x and y with respect to t, then divide dy/dt by dx/dt. This works only where dx/dt ≠ 0
🎡 The tangent to a cycloid
①When a wheel of radius 1 has turned through angle t, a point on its rim sits at x = t − sin t, y = 1 − cos t. The picture uses the same scale on both axes, so the red point always lies on the blue wheel
②Since dx/dt = 1 − cos t and dy/dt = sin t, the tangent slope is sin t / (1 − cos t)
③That line is the same as the line joining the red point to the top of the wheel at (t, 2), which is why the red tangent keeps passing through the white circle wherever you move the slider
④Where 1 − cos t = 0, as at t = 2π, the point touches the ground at a cusp, and near it the slope grows without bound in size
Logarithmic Differentiation
Logarithmic differentiation
y = xx → ln y = x ln x → y'/y = ln x + 1 → y' = xx(ln x + 1)
For x > 0, taking the natural log of both sides brings the x in the exponent down as a factor, so the expression can be differentiated. Differentiating ln y on the left gives y′/y by the chain rule
💡 When to use it
①When the variable is in both the base and the exponent, as in xˣ. Neither the power rule nor the exponential rule applies as it stands
②When many factors are multiplied and divided together. Taking ln turns products into sums and quotients into differences
③After differentiating, multiply both sides by y to leave y′ on its own
Worked Examples
Example 1
Differentiate y = (2x+1)5.
1
Let the inside be u = 2x + 1, so y = u5.
y = u5, u = 2x + 1
2
Multiply the outer derivative 5u4 by the inner derivative (2x + 1)′ = 2.
y′ = 5(2x+1)4 · 2 = 10(2x+1)4
▸ y′ = 10(2x+1)4
Leaving out the inner derivative 2 gives the common wrong answer 5(2x+1)4.
Example 2
Find the slope of the tangent to the circle x2 + y2 = 25 at the point (3, 4).
1
Differentiate both sides with respect to x. The term y2 becomes 2y · y′.
2x + 2y·y' = 0 ⇒ y' = -xy
2
Substitute the point (3, 4).
y'|(3,4) = -34
▸ -34
The radius to (3, 4) has slope 4/3, and multiplying it by the tangent slope −3/4 gives −1. The fact that a tangent is perpendicular to the radius leads to the same answer.
Summary
Three methods
Chain: f'·g' Implicit: differentiate, solve for y' Parametric: g'(t)f'(t)
The chain rule, implicit differentiation and parametric differentiation all come down to multiplying and dividing small changes
KICE mock exam type, adapted
For the curve given parametrically by x = t2 + 1, y = t3 − t, what is dydx at t = 2?
①94
②52
③114
④3
⑤134
▸ ③ 114
1
Differentiating with respect to t gives dx/dt = 2t and dy/dt = 3t2 − 1.
dydx = 3t2 - 12t
2
At t = 2 the numerator is 11 and the denominator is 4.
3(2)2 - 12(2) = 114
🎯 Exam Points
①The chain rule comes up most often, and the factor students drop is the derivative of the inside
②In implicit differentiation, every term containing y picks up a factor y′
③For parametric curves dy/dx = (dy/dt)/(dx/dt), and points where dx/dt = 0 are checked separately
④For xˣ, take ln of both sides and then differentiate
⑤Problems that need the chain rule and implicit differentiation together are common