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

y=3x8dydx=3 \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 x in place:

y+3x+8=0dydx+3+0=0dydx=3 \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 x2+x+y2=15 x^2 + x + y^2 = 15 what is dydx \frac{dy}{dx} at the point (2,3) ( 2, 3) ?

Since x2+x+y2=15 x^2 + x + y^2 = 15 , differentiating, we have:

2x+1+2y(dydx)=0dydx=2x+12y \begin{aligned} 2x+1+2y\left(\frac{dy}{dx}\right)&=0 \\ \frac{dy}{dx}&=\frac{2x+1}{2y} \end{aligned}

Thus dydx \frac{dy}{dx} at the point (2,3) ( 2, 3) is 2(2)+12(3)=56 \frac{2(2)+1}{2(3)}=\frac{5}{6} . _\square

 

If y6exy=x y^6 - e^{xy} = x what is dydx \frac{dy}{dx} ?

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

6y5(dydx)exy(ddx(xy))=16y5(dydx)exy(y+xdydx)=1 \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 dydx \frac{dy}{dx} on the left:

6y5(dydx)xexy(dydx)=1+yexy(dydx)(6y5xexy)=1+yexydydx=1+yexy6y5xexy \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
5 years, 6 months ago

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nice

Wesllen Brendo - 5 years, 4 months ago

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