Implicit differentiation of a function of two or more variables
Suppose the function
$z=f\left(x,y\right)$ defines
$y$ implicitly as a function
$y=g\left(x\right)$ of
$x$ via the equation
$f\left(x,y\right)=0.$ Then
If the equation
$f\left(x,y,z\right)=0$ defines
$z$ implicitly as a differentiable function of
$x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y,$ then
[link] is a direct consequence of
[link] . In particular, if we assume that
$y$ is defined implicitly as a function of
$x$ via the equation
$f\left(x,y\right)=0,$ we can apply the chain rule to find
$dy\text{/}dx\text{:}$
Solving this equation for
$dy\text{/}dx$ gives
[link] .
[link] can be derived in a similar fashion.
Let’s now return to the problem that we started before the previous theorem. Using
[link] and the function
$f\left(x,y\right)={x}^{2}+3{y}^{2}+4y-4,$ we obtain
which is the same result obtained by the earlier use of implicit differentiation.
Implicit differentiation by partial derivatives
Calculate
$dy\text{/}dx$ if
$y$ is defined implicitly as a function of
$x$ via the equation
$3{x}^{2}-2xy+{y}^{2}+4x-6y-11=0.$ What is the equation of the tangent line to the graph of this curve at point
$\left(2,1\right)?$
Calculate
$\partial z\text{/}\partial x$ and
$\partial z\text{/}\partial y,$ given
${x}^{2}{e}^{y}-yz{e}^{x}=0.$
Set
$f(x,y)=3{x}^{2}-2xy+{y}^{2}+4x-6y-11=0,$ then calculate
${f}_{x}$ and
${f}_{y}\text{:}$$\begin{array}{c}{f}_{x}=6x-2y+4\hfill \\ {f}_{y}=\mathrm{-2}x+2y-6.\hfill \end{array}$ The derivative is given by
Find
$dy\text{/}dx$ if
$y$ is defined implicitly as a function of
$x$ by the equation
${x}^{2}+xy-{y}^{2}+7x-3y-26=0.$ What is the equation of the tangent line to the graph of this curve at point
$\left(3,\mathrm{-2}\right)?$
$\frac{dy}{dx}={\frac{2x+y+7}{2y-x+3}|}_{\left(3,\mathrm{-2}\right)}=\frac{2\left(3\right)+\left(\mathrm{-2}\right)+7}{2\left(\mathrm{-2}\right)-\left(3\right)+3}=-\frac{11}{4}$ Equation of the tangent line:
$y=-\frac{11}{4}x+\frac{25}{4}$
The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables.
Tree diagrams are useful for deriving formulas for the chain rule for functions of more than one variable, where each independent variable also depends on other variables.
Key equations
Chain rule, one independent variable $\frac{dz}{dt}=\frac{\partial z}{\partial x}\xb7\frac{dx}{dt}+\frac{\partial z}{\partial y}\xb7\frac{dy}{dt}$
For the following exercises, use the information provided to solve the problem.
Let
$w\left(x,y,z\right)=xy\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}z,$ where
$x=t,y={t}^{2},$ and
$z=\text{arcsin}\phantom{\rule{0.2em}{0ex}}t.$ Find
$\frac{dw}{dt}.$
If
$w=x{y}^{2},x=5\phantom{\rule{0.2em}{0ex}}\text{cos}(2t),$ and
$y=5\phantom{\rule{0.2em}{0ex}}\text{sin}(2t),$ find
$\frac{\partial w}{\partial t}.$
If
$f(x,y)=xy,x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,$ and
$y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ find
$\frac{\partial f}{\partial r}$ and express the answer in terms of
$r$ and
$\theta .$
Suppose
$f(x,y)=x+y,u={e}^{x}\text{sin}\phantom{\rule{0.2em}{0ex}}y,x={t}^{2},$ and
$y=\pi t,$ where
$x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta $ and
$y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$ Find
$\frac{\partial f}{\partial \theta}.$
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can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?