MTH2010 Chap.4 Chain Rule, Implicit Differentiation and the Gradient
Chain Rule, Implicit Differentiation and the Gradient
Week 4 extends differentiation to compositions and to functions defined implicitly, then introduces the gradient, the single most useful vector in the unit. The multivariable chain rule says that when z depends on x and y, and each of those depends on further variables s and t, the derivative of z with respect to s is a sum of products, one for every route from z down to s in a dependency tree.
The derivatives of the outer function are evaluated at the inner point, a detail that the homework sheets test directly. A related trap separates the total derivative of a function along a curve from a partial derivative that holds the other variable fixed.
Implicit differentiation follows from the chain rule: if F(x, y) = 0 defines y as a function of x, then dy/dx = -F_x / F_y, and the implicit function theorem guarantees that such a function exists near a point when F has continuous partial derivatives, the point satisfies the equation, and F_y is non-zero there.
The same formula gives partial derivatives of z on a surface F(x, y, z) = 0. The directional derivative measures the rate of change along a unit vector and equals that vector dotted with the gradient. As a result the gradient points in the direction of fastest increase, its length is that maximum rate, and it is perpendicular to level curves and level surfaces.
Those facts produce tangent planes to level surfaces and tangent lines to curves of intersection.
What this chapter covers
- 01
Chain rule: one product per route in the dependency tree
- 02
Evaluation points: outer partials at the inner point, inner partials at (s, t)
- 03
Total versus partial derivatives: the homework's fake proof
- 04
Implicit function theorem: three conditions to check before using the formula
- 05
Implicit derivatives: dy/dx = -F_x / F_y and z_x = -F_x / F_z
- 06
Directional derivatives: unit vector dotted with the gradient
- 07
Steepest ascent and descent: direction and maximum rate
- 08
Gradient as normal: tangent planes to level surfaces and curves of intersection
Worked example · free
Directional derivative in three variables
- 1Gradient: (2xy, x² + z, y) = (4, 1 - 1, 2) = (4, 0, 2) at the point.
- 1Unit direction: (2, 1, 2) has length 3, so u = (2, 1, 2)/3.
- 1Directional derivative: (4*2 + 0*1 + 2*2)/3 = 12/3 = 4.
- 1Maximum rate: |(4, 0, 2)| = sqrt(20) = 2 sqrt(5), about 4.47, in the direction (4, 0, 2).
Key terms
- Dependency tree
- A diagram of which variables depend on which, used to list the products in a chain rule.
- Implicit function theorem
- A result guaranteeing that an equation F(x, y) = 0 defines y as a differentiable function of x near a point where F_y is non-zero.
- Directional derivative
- The instantaneous rate of change of a function as the input moves along a given unit vector.
- Steepest ascent direction
- The direction of the gradient, along which a function increases fastest.
- Level surface normal
- The gradient of F at a point, which is perpendicular to the level surface F = c through that point.
Chain Rule, Implicit Differentiation and the Gradient FAQ
Where do I evaluate the partial derivatives in the chain rule?
Evaluate the outer function's partials at the inner point, meaning the values of x and y produced by the given s and t. Evaluate the inner derivatives at s and t themselves, then multiply along each route and add.
What must I check before using implicit differentiation?
Confirm that the point satisfies the equation, that the partial derivatives of F are continuous nearby, and that the partial derivative with respect to the dependent variable is non-zero there. Then the implicit function theorem applies.
Why does the gradient point in the direction of fastest increase?
The directional derivative is the dot product of a unit vector with the gradient, which is largest when the unit vector points the same way as the gradient. The largest value equals the gradient's length.
How do I find the tangent line to the intersection of two surfaces?
Compute the gradient of each surface at the point. Both are normal to the curve of intersection, so their cross product points along it, and the tangent line runs through the point in that direction.
Exam move
Draw a dependency tree for every chain-rule problem, even simple ones, and label where each factor is evaluated. Check numerical answers by direct substitution when the composite is short. For implicit differentiation, write the three conditions of the implicit function theorem as a checklist and tick them before differentiating.
For gradients, practise the same three outputs on one function: a directional derivative with a non-unit direction, the steepest direction with its rate, and a tangent plane to a level surface. Then try one question that asks for a direction giving a fraction of the maximum rate, solved with the cosine of the angle.
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