4.5 Implicit differentiation

Where did the $y'$ come from?

(a) and (b)

We use the power rule (a) and the chain rule (b): the power rule tells us to “bring down the 2", and the chain rule tells us to multiply by $y'$. There is no need for the quotient rule here, as there are no quotients. Exponential functions have the form $(\text{constant})^{\text{function}}$, but our function has the form $(\text{function})^{\text{constant}}$, so we did not use (d).

The three points to look at are $(0,-4)$, $(0,0)$, and $(0,4)$. What does the slope of the tangent line look like there?

At $(0,4)$ and $(0,-4)$, $\ds\diff{y}{x}$ is 0; at $(0,0)$, $\ds\diff{y}{x}$ does not exist.

At $(0,4)$ and $(0,-4)$, the curve looks to be horizontal, if you zoom in: a tangent line here would have derivative zero. At the origin, the curve looks like its tangent line is vertical, so $\ds\diff{y}{x}$ does not exist.

A function must pass the vertical line test: one input cannot result in two different outputs.

(a) no (b) no
$\ds\diff{y}{x}=-\dfrac{x}{y}$. It is not possible to write $\ds\diff{y}{x}$ as a function of $x$, because (as stated in (b)) one value of $x$ may give two values of $\ds\diff{y}{x}$. For instance, when $x=\pi/4$, at the point $\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)$ the circle has slope $\ds\diff{y}{x}=-1$, while at the point $\left(\dfrac{\pi}{4},\dfrac{-1}{\sqrt{2}}\right)$ the circle has slope $\ds\diff{y}{x}=1$.

(a) No. A function must pass the vertical line test: one input cannot result in two (or more) outputs. Since one value of $x$ sometimes corresponds to two values of $y$ (for example, when $x=\pi/4$, $y$ is $\pm 1/\sqrt{2}$), there is no function $f(x)$ so that $y=f(x)$ captures every point on the circle.

Remark: $y=\pm\sqrt{1-x^2}$ does capture every point on the unit circle. However, since one input $x$ sometimes results in two outputs $y$, this expression is not a function.

(b) No, for the same reasons as (a). If $f'(x)$ is a function, then it can give at most one slope corresponding to one value of $x$. Since one value of $x$ can correspond to two points on the circle with different slopes, $f'(x)$ cannot give the slope of every point on the circle. For example, fix any $0<a<1$. There are two points on the circle with $x$-coordinate equal to $a$. At the upper one, the slope is strictly negative. At the lower one, the slope is strictly positive.

(c) We differentiate:

and solve for $\ds\diff{y}{x}$

But there is a $y$ in the right-hand side of this equation, and it's not clear how to get it out. Our answer in (b) tells us that, actually, we can't get it out, if we want the right-hand side to be a function of $x$. The derivative cannot be expressed as a function of $x$, because one value of $x$ corresponds to multiple points on the circle.

Remark: since $y=\pm\sqrt{1-x^2}$, we could try writing

but this is not a function of $x$. Again, in a function, one input leads to at most one output, but here one value of $x$ will usually lead to two values of $\diff{y}{x}$.

The problem isn't with any of the algebra.

The derivative $\ds\diff{y}{x}$ is $\dfrac{11}{4}$ only at the point $(1,3)$: it is not constantly $\dfrac{11}{4}$, so it is wrong to differentiate the constant $\dfrac{11}{4}$ to find $\ds\ddiff{2}{y}{x}$. Below is a correct solution.

Plugging in $x=1$, $y=3$:

Differentiating the equation $-28x+2y+2xy'+2yy'=0$:

At the point $(1,3)$, $y'=\dfrac{11}{4}$. Plugging in:

The derivative $\ds\diff{y}{x}$ is $\dfrac{11}{4}$ only at the point $(1,3)$: it is not constantly $\dfrac{11}{4}$, so it is wrong to differentiate the constant $\dfrac{11}{4}$ to find $\ds\ddiff{2}{y}{x}$. Below is a correct solution.

Plugging in $x=1$, $y=3$:

Differentiating the equation $-28x+2y+2xy'+2yy'=0$:

At the point $(1,3)$, $y'=\dfrac{11}{4}$. Plugging in:

Remember that $y$ is a function of $x$. Use implicit differentiation, then collect all the terms containing $\ds\diff{y}{x}$ on one side of the equation to solve for $\ds\diff{y}{x}$.

$\ds\diff{y}{x}=-\dfrac{e^x+y}{e^y+x}$

Remember that $y$ is a function of $x$. We begin with implicit differentiation.

Now, we solve for $\ds\diff{y}{x}$.

Differentiate implicitly, then solve for $y'$.

$\ds\diff{y}{x} = \dfrac{y^2+1}{e^y-2xy}$

Differentiate both sides of the equation with respect to $x$:

Now, get the derivative on one side and solve

Remember that $y$ is a function of $x$. You can determine explicitly the values of $x$ for which $y(x)=1$.

At $(x,y)=(4,1)$, $y' = -\dfrac{1}{\pi + 1}$. At $(x,y)=(-4,1)$, $y' = \dfrac{1}{\pi-1}$.

You don't need to solve for $y''$ in general–only when $x=y=0$. To do this, you also need to find $y'$ at the point $(0,0)$.

$-4$

We differentiate implicitly. For ease of notation, we write $y'$ for $\ds\diff{y}{x}$.

We're interested in $y''$, so we implicitly differentiate again.

We want to know what $y''$ is when $x=y=0$. Plugging these in yields the following:

So, we need to know what $y'$ is when $x=y=0$. We can get this from the equation $2x+1+y'=\cos(xy)(y+xy')$, which becomes $1+y'=0$ when $x=y=0$. So, at the origin, $y'=-1$, and

Remark: a common mistake is to stop at the equation $2x+1+y'=\cos(xy)(y+xy')$, plug in $x=y=0$, find $y'=-1$, and decide $y''=\ds\diff{}{x}\{-1\}=0$. This is due to a slight sloppiness in the usual notation. When we wrote $y'=1$, what we meant is that at the point $(0,0)$, $\ds\diff{y}{x}=-1$. More properly written: $\left.\ds\diff{y}{x}\right|_{x=0,~y=0}=-1$. This is not the same as saying $y'=1$ everywhere (in which case, indeed, $y''$ would be 0 everywhere).

$-\ds\frac{2x\sin(x^2+y)+3x^2}{4y^3+\sin(x^2+y)}$

Differentiate the equation and solve:

Plug in $y=0$ at a strategic point in your work to simplify your computation.

At $(x,y)=(1,0)$, $y' = -6$, and at $(x,y)=(-5,0)$, $y' = \frac{6}{25}$.

Use implicit differentiation.

$\ds\ddiff{2}{y}{x}=\dfrac{-1}{y^3}$

We use implicit differentiation, twice.

So, we need an expression for $y'$. We use the equation $2x+2yy'=0$ to conclude $y'=-\dfrac{x}{y}$:

$\diff{y}{x} = \dfrac{\cos(x+y)-2x}{2y-\cos(x+y)}$

Differentiate the equation and solve:

Plug in $y = 0$ at a strategic point in your work to simplify your computation.

At $(x,y)=(2,0)$ we have $y' = -\frac{3}{2}$, and at $(x,y)=(-4,0)$ we have $y' = -\frac{3}{4}$.

If the tangent line has slope $y'$, and it is parallel to $y=x$, then $y'=1$.

$\left(\dfrac{\sqrt{3}}{2},\dfrac{-1}{2\sqrt{3}}\right)$, $\left(\dfrac{-\sqrt{3}}{2},\dfrac{1}{2\sqrt{3}}\right)$

The question asks at which points on the ellipse $\ds\diff{y}x{}=1$. So, we begin by differentiating, implicitly:

We could solve for $\ds\diff{y}{x}$ at this point, but it's not necessary. We want to know when $\ds\diff{y}{x}$ is equal to one:

That is, $\ds\diff{y}{x}=1$ at those points along the ellipse where $x=-3y$. We plug this into the equation of the ellipse to find the coordinates of these points.

So, the points along the ellipse where the tangent line is parallel to the line $y=x$ occur when $y=\dfrac{1}{2\sqrt{3}}$ and $x=-3y$, and when $y=\dfrac{-1}{2\sqrt{3}}$ and $x=-3y$. That is, the points $\left(\dfrac{-\sqrt{3}}{2},\dfrac{1}{2\sqrt{3}}\right)$ and $\left(\dfrac{\sqrt{3}}{2},\dfrac{-1}{2\sqrt{3}}\right)$.

You don't need to solve for $y'$ in general: only at a single point.

$-\dfrac{28}{3}$

First, we differentiate implicitly with respect to $x$.

Now, we plug in $x=1$, $y=4$, and solve for $\ds\diff{y}{x}$:

After you differentiate implicitly, get all the terms containing $y'$ onto one side so you can solve for $y'$.

$\ds\diff{y}{x}=-\dfrac{2xy^2+\sin y}{2x^2y+x\cos y}$

Implicitly differentiating $x^2y(x)^2+x\sin(y(x))=4$ with respect to $x$ gives

Then we gather the terms containing $y'$ on one side, so we can solve for $y'$:

Recall $\ds\diff{}{x}\log x =\ds\frac{1}{x}$.

$f''(x)=\dfrac{1}{x}$

You don't need to solve for $\diff{y}{x}$ for all values of $x$–only when $y=0$.

At $(x,y)=(2,0)$, $y' = -2$. At $(x,y)=(-2,0)$, $y' = 2$.

$x=0$, $x=1$, $x=-1$

The slope of the tangent line is, of course, given by the derivative, so let's start by finding $\diff{y}{x}$ of both shapes.

For the circle, we differentiate implicitly

and solve for $\ds\diff{y}{x}$

For the ellipse, we also differentiate implicitly:

and solve for $\ds\diff{y}{x}$

What we want is a value of $x$ where both derivatives are equal. However, they might have different values of $y$, so let's let $y_1$ be the $y$-values associated with $x$ on the circle, and let $y_2$ be the $y$-values associated with $x$ on the ellipse. That is, $x^2+ y_1^2=1$ and $x^2+3y_2^2=1$. For the slopes at $(x,y_1)$ on the circle and $(x,y_2)$ on the ellipse to be equal, we need: