At the end of the course, each student should be able to…
- C06: ChainRule. Apply the multivariable Chain Rule to compute derivatives and find normal vectors.
C06. Chain Rule
- The single-variable Chain Rule \(\frac{d}{dt}[f(u(x))]=f’(u(x))u’(x)\)
may be generalized for multiple variable functions.
- If \(f(P)\) is a function of multiple variables and \(\vect{r}(t)\) is a vector function of equal dimension, then \(\frac{d}{dt}[f(\vect r(t))]=\nabla f(\vect r(t))\cdot\vect r’(t)\).
- An immediate result is that \(\nabla f(P_0)\) is normal to the level curve of \(f\) passing through \(P_0\).
- Thus \(\<f_x(P_0),f_y(P_0),-1\>\) is normal to the surface \(z=f(x,y)\) at \(P_0\), and \(z=f(P_0)+f_x(P_0)(x-x_0)+f_y(P_0)(y-y_0)\) is the tangent plane to the surface \(z=f(x,y)\) at \(P_0\).
- The total derivative \(\frac{df}{dx}\) describes the rate of change of
\(f\) with respect to \(x\) when the other variables of \(f\)
are dependent on \(x\) as well.
- The chain rule shows us \(\frac{df}{dx}=\nabla f\cdot\frac{d\vect r}{dx}\).
- For three variables: \(\frac{df}{dx}=\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}+ \frac{\partial f}{\partial z}\frac{dz}{dx}\).
- Thus if \(f(x,y)=c\) defines \(y\) as a differentiable function of \(x\), then \(\frac{dy}{dx}=-\frac{f_x}{f_y}\).
Textbook References
- University Calculus: Early Transcendentals (3rd Ed)
- 13.4 (exercises 1-6, 25-32)
- 13.6 (exercises 1-12)