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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:

\[ \begin{align} y &= -3x - 8 \\ \frac{dy}{dx} &= -3 \end{align} \]

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

\[ \begin{align} y + 3x + 8 &= 0 \\ \frac{dy}{dx} + 3 + 0 &=0 \\ \frac{dy}{dx} &= -3 \end{align} \]

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:

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

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:

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

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

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

Note by Arron Kau
3 years, 10 months ago

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nice

Wesllen Brendo - 3 years, 7 months ago

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