# Implicit Differentiation

## Definition

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.

## Technique

### 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}

Note by Arron Kau
7 years, 2 months ago

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nice

- 7 years ago