Implicit Differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables.

For example, if $y + 3x + 8 = 0$, we could solve for $y$ and then differentiate:

$\begin{aligned} y &= -3x - 8 \\ \frac{dy}{dx} &= -3 \end{aligned}$

However, we could also simply take the derivative of each term with respect to $x$ in place:

$\begin{aligned} y + 3x + 8 &= 0 \\ \frac{dy}{dx} + 3 + 0 &=0 \\ \frac{dy}{dx} &= -3 \end{aligned}$

The second approach is known as *implicit differentiation*.

## Given $x^2 + x + y^2 = 15$ what is $\frac{dy}{dx}$ at the point $( 2, 3)$?

Since $x^2 + x + y^2 = 15$, differentiating, we have:

$\begin{aligned} 2x+1+2y\left(\frac{dy}{dx}\right)&=0 \\ \frac{dy}{dx}&=\frac{2x+1}{2y} \end{aligned}$

Thus $\frac{dy}{dx}$ at the point $( 2, 3)$ is $\frac{2(2)+1}{2(3)}=\frac{5}{6}$. $_\square$

## If $y^6 - e^{xy} = x$ what is $\frac{dy}{dx}$?

Taking the derivative of every term with respect to $x$ gives us:

$\begin{aligned} 6y^5\left( \frac{dy}{dx} \right) - e^{xy} \left(\frac{d}{dx} (xy)\right)&=1 \\ 6y^5\left( \frac{dy}{dx} \right) - e^{xy} \left( y + x\frac{dy}{dx} \right)&=1 \end{aligned}$

Now, we can isolate all of the $\frac{dy}{dx}$ on the left:

$\begin{aligned} 6y^5\left( \frac{dy}{dx} \right) - xe^{xy}\left(\frac{dy}{dx}\right) &=1 + ye^{xy} \\ \left( \frac{dy}{dx} \right) \left(6y^5 - xe^{xy} \right)&=1 + ye^{xy} \\ \frac{dy}{dx} &= \frac{ 1 + ye^{xy} }{ 6y^5 - xe^{xy} } _\square \end{aligned}$

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestnice

Log in to reply