Implicit Differentiation#

Introduction

For each of the following problems:

Find $y^{\prime}$ by solving the equation for $y$ and differentiating directly.
Find $y^{\prime}$ by implicit differentiation.
Check the derivatives obtained through both methods are the same.

$\frac{x}{y^3} = 1$
$x^2 + y^3 = 4$

Applications

Use implicit differentiation to solve the following problems:

$2 \cdot y^3 + 4 \cdot x^2 - y = x^6$
$7 \cdot y^2 + \sin(3x) = 12 - y^4$
$e^x - \sin(y) = x$
$\cos(x^2 + 2y) + x \cdot e^{y^2} = 1$
$\tan(x^2 \cdot y^4) = 3x + y^2$

A Pointless Problem

Important

Take time to appreciate the pun here. If you don’t get it yet, you will.

Consider the function $f(x,y)$ ,

$x^2 + y^2 = 9$

Plot this function in the x-y plane. (Desmos is good for this). What kind of graph is this?

Important

Do you get it now?!

Use implicit differentiation to find $\frac{dy}{dx}$ . Express the answer as a function of $x$ only.
Find the equation of the tangent line at the points $(\pm \frac{3 \cdot \sqrt{2}}{2}, \pm \frac{3 \cdot \sqrt{2}}{2})$
Plot the tangent lines on top of the graph you created in part a.
Find the points $(x,y)$ where the tangent lines found in part c intersect.
Plot the points found in part e on top of the graphs you created in part a and part d.
What is the area of the quadrilateral formed by the tangent lines in part c?

Natural Log Derivative

The formula for the derivative of the natural log, $ln(x)$ , can be derived with implicit differentiation. To do so, recall the differentiaion rule for exponential functions,

Then define y as,

$y = \ln(x)$

Use implicit differentiation to derive the formula for,

$\frac{d}{dx}(ln(x)) = \frac{1}{x}$

Hint

Solve for $x$ and then apply the Chain Rule.

Inverse Trigonometric Derivatives

In class we used implicit differentiation to derive,

$\frac{d}{dx}( \arcsin(x) ) = \frac{1}{\sqrt{1-x^2}}$

Using a similar process, find the derivatives of the following inverse trigonometric functions,

$f(x) = \arccos(x)$
$f(x) = \arctan(x)$

Hint

Remember to draw a diagram of the unit circle. Express x and y in terms of lengths and angles!

2005, Free Response Form B, #5

Consider the curve given by,

$y ^2 = 2 + xy$

Show that

$\frac{dy}{dx} = \frac{y}{2y -x}$

Find all points $(x,y)$ on the curve where the line tangent to the curve has slope $\frac{1}{2}$ .
Show that there are no points $(x,y)$ on the curve where the line tangent to the curve is horizontal.

d. Let x and y be functions of time t that are related by the equation $y^2 = 2 + xy$ . At time $t = 5$ , the value of $y$ is 3 and $\frac{dy}{dt} = 6$ . Find the value of $\frac{dx}{dt}$ at time $t = 5$ .

2023, Free Response, #6

Consider the curve given by the equation,

$6xy = 2 + y^3$

Show that,

$\frac{dy}{dx} = \frac{2y}{y^2 - 2x}$

Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.
Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.
A particle is moving along the curve. At the instance when the particle is at the point $(\frac{1}{2}, -2)$ , its horizontal position is increasing at a rate of $\frac{dx}{dt}=\frac{2}{3}$ units per second. What is the value of $\frac{dy}{dt}$ , the rate of change of the particle’s vertical position, at that instant?

2015, Free Response, #6

Consider the curve given by the equation $y^3 - xy = 2$ . It can be shown that $\frac{dy}{dx} = \frac{y}{3y^2 - x}$ .

Write an equation for the line tangent to the curve at the point $(-1, 1)$ .
Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.
Evaluate $\frac{d^2 y}{dx^2}$ at the point on the curve where $x = -1$ and $y = 1$ .