Collect the derivative terms
For $x^2+y^2=25$:
$$2x+2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{x}{y}.$$
At $(3,4)$, the tangent gradient is $-3/4$.
Implicit differentiation
Differentiate relations containing both variables and use the resulting derivative to study tangents and curvature.
By the end of this lesson
- differentiate $y$-terms using the chain rule
- solve an implicit derivative for $dy/dx$
- find tangent gradients and a second derivative where suitable
Why this matters. Many curves cannot be written conveniently as one explicit function. Implicit differentiation works directly with their defining relation.
Treat y as a function of x
When differentiating with respect to $x$, treat $y$ as $y(x)$. The chain rule supplies $dy/dx$:
$\dfrac d{dx}(y)=\dfrac{dy}{dx}$$\dfrac d{dx}(y^2)=2y\dfrac{dy}{dx}$$\dfrac d{dx}(\sin y)=\cos y\dfrac{dy}{dx}$
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.
Show explanation
$y$ is a function of $x$, so the chain rule adds $dy/dx$.
Example 1
Differentiate a circle
Use product and chain rules together
If an expression contains $xy$, then both factors vary with $x$:
$$\frac d{dx}(xy)=x\frac{dy}{dx}+y.$$
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.
Show explanation
The product $xy$ gives $x(dy/dx)+y$, and $y^2$ gives $2y(dy/dx)$.
Example 2
Find a tangent
A mixed relation
For $x^2+xy+y^2=7$, collect derivative terms:
$$\left(x+2y\right)\frac{dy}{dx}=-(2x+y)$$
$$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$
At $(1,2)$, the gradient is $-4/5$, so the tangent is
$$y-2=-\frac45(x-1).$$
Differentiate again with care
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 $x$.
Show explanation
Differentiate $yy'$ with the product rule: $(y')^2+yy''$.
Example 3
Find curvature information
Second derivative on a circle
From $2x+2yy'=0$, differentiate again:
$$2+2(y')^2+2yy''=0\Rightarrow y''=-\frac{1+(y')^2}{y}.$$
At $(3,4)$, $y'=-3/4$, hence
$$y''=-\frac{1+9/16}{4}=-\frac{25}{64}.$$
The negative sign indicates the upper branch is concave down there.
In the exam
Quick check
Five questions to check the main decisions from this lesson.
Your score is not saved.
Apply the chain rule to $y(x)^4$.
Show explanation
Apply the chain rule to $y(x)^4$.
Use the product rule.
Show explanation
Use the product rule.
A horizontal line has zero gradient.
Show explanation
A horizontal line has zero gradient.
$y'=-x/y=0$.
Show explanation
$y'=-x/y=0$.
$y$ and its derivative are both functions of $x$.
Show explanation
$y$ and its derivative are both functions of $x$.
Lesson Rewards
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