Differentiate relations containing both variables and use the resulting derivative to study tangents and curvature.
On $x^2+y^2=25$, moving along the circle changes $y$ as $x$ changes, even though the equation has not isolated $y$. We can differentiate the relation directly, following one branch of the curve.
Check the logic
At the rightmost point of the circle, is the tangent horizontal or vertical?
The radius there is horizontal, so its perpendicular tangent is vertical; dy/dx is undefined.
Move the point to angle 0°, then 90°. Compare the tangent directions and explain where −x/y cannot be used.
On x²+y²=25, differentiating gives 2x+2y dy/dx=0. At (3,4) the slope is −3/4. At y=0 the tangent is vertical and dy/dx is undefined.
Differentiating $x^2$ with respect to $x$ gives $2x$. For $y^2$, the outer square changes with $y$, and $y$ itself changes with $x$. The chain rule accounts for both:
Here $dy/dx$ is the local tangent slope, not an extra variable added to balance the equation.
Write $y=y(x)$. Then $\frac{d}{dx}[y(x)]^2=2y(x)y'(x)$, just as differentiating $[g(x)]^2$ gives $2g(x)g'(x)$. This describes a differentiable local branch; at the circle’s left and right extremes, the tangent is vertical and a finite $dy/dx$ does not exist.
The same rule applies to other outer functions:
Checkpoint
Correct. $y$ is a function of $x$, so the chain rule adds $dy/dx$.
Not quite. Differentiating as though $y$ were the variable of differentiation omits the necessary chain factor.
$y$ is a function of $x$, so the chain rule adds $dy/dx$.
Example 1
Collect the derivative terms
For $x^2+y^2=25$:
Check the logic
Identify the missing factor: a student differentiates y² as 2y. What must multiply 2y?
y changes with x, so the chain rule contributes dy/dx.
Look for that factor before collecting the derivative terms.
At $(3,4)$, the tangent gradient is $-3/4$.
If an expression contains $xy$, then both factors vary with $x$:
Checkpoint
Correct. The product $xy$ gives $x(dy/dx)+y$, and $y^2$ gives $2y(dy/dx)$.
Not quite. Both the product rule and the chain rule are required.
The product $xy$ gives $x(dy/dx)+y$, and $y^2$ gives $2y(dy/dx)$.
Example 2
A mixed relation
For $x^2+xy+y^2=7$, collect derivative terms:
At $(1,2)$, the gradient is $-4/5$, so the tangent is
A second derivative may be found by differentiating the first derivative relation, then substituting the known value of $dy/dx$.
Checkpoint
Correct. Differentiate $yy'$ with the product rule: $(y')^2+yy''$.
Not quite. Both $y$ and $y'$ vary with $x$.
Differentiate $yy'$ with the product rule: $(y')^2+yy''$.
Example 3
Second derivative on a circle
From $2x+2yy'=0$, differentiate again:
At $(3,4)$, $y'=-3/4$, hence
The negative sign indicates the upper branch is concave down there.
Your score is not saved.
Apply the chain rule to $y(x)^4$.
Apply the chain rule to $y(x)^4$.
Use the product rule.
Use the product rule.
A horizontal line has zero gradient.
A horizontal line has zero gradient.
$y'=-x/y=0$.
$y'=-x/y=0$.
$y$ and its derivative are both functions of $x$.
$y$ and its derivative are both functions of $x$.
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