Before the midterm

Forgetting dy/dx on a term in y

The habit: y depends on x, so the chain rule brings out a factor of dy/dx from everything in y that is differentiated.

10 practice questions are built to catch it, in 3 topics.

See it fail in a figure: The slip: no y′, in the implicit differentiation and inverse trigonometric functions lesson

Practise against it: 6 questions

What it looks like

A question from implicit differentiation and inverse trigonometric functions, with the answer this slip gives beside the right one.

Introductory1 mark · 2 min

In implicit differentiation, $y$ stands for an unknown function of $x$ . Which expression equals $\dfrac{d}{dx}\bigl(y^{3}\bigr)$ ?

With the slip: $\displaystyle 3y^{2}$
Worked through: $\displaystyle 3y^{2}\,\dfrac{dy}{dx}$

3y² is the derivative with respect to y. Here the variable is x, and y depends on x, so the chain rule adds a factor.

See the worked solution

Where it comes up

Implicit differentiation and inverse trigonometric functions8 questions
Linear approximation1 question
Functions of two variables and partial derivatives1 question

Practise against it

Each of these has an answer this slip gives. Work them on paper: if you fall into it, the verdict names it and the question comes back in review.

01
Given $\sin(y) + x^{2} = y$ , which expression is $\dfrac{dy}{dx}$ ?
Implicit differentiation and inverse trigonometric functionsStandard
02
The point $(2, 1)$ lies on the curve below. Find $\dfrac{dy}{dx}$ at that point. $\displaystyle x^{3} + \sin(\pi y) + y = 9$
Implicit differentiation and inverse trigonometric functionsStandard
03
Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ for the curve below. $\displaystyle xy^{2} + e^{y} = 2x$
Implicit differentiation and inverse trigonometric functionsStandard
04
The curve below passes through exactly one point with $y = 0$ . Find $\dfrac{dy}{dx}$ at that point. $\displaystyle x\,e^{y} + 2x^{2}\sin y = 1$
Implicit differentiation and inverse trigonometric functionsStandard
05
For the circle below, find $\dfrac{d^{2}y}{dx^{2}}$ at a point where $y > 0$ , in terms of $x$ and $y$ . $\displaystyle x^{2} + y^{2} = 25$
Implicit differentiation and inverse trigonometric functionsChallenging
06
The point $(-1, 2)$ lies on the curve below. Find $\dfrac{d^{2}y}{dx^{2}}$ at that point. $\displaystyle x^{3} + y^{3} = 7$
Implicit differentiation and inverse trigonometric functionsChallenging