Differentiation

Lesson

It can happen that a variable $y$`y` is known to be a function of $x$`x` but we do not have an explicit definition of the function. Instead, we have an equation that is satisfied by $x$`x` and a function $y$`y` of $x$`x`. Often, there may be more than one function $y$`y` that satisfies the equation.

For example, the equation $x^2+y^2=1$`x`2+`y`2=1 is satisfied by the points $(x,y)$(`x`,`y`) that belong to the unit circle. In this case, it is possible to rearrange the equation into a form that will serve as a function definition. In particular, $y=\sqrt{1-x^2}$`y`=√1−`x`2 with the domain $-1\le x\le1$−1≤`x`≤1. But note that $y=-\sqrt{1-x^2}$`y`=−√1−`x`2 also satisfies the equation so that, in this case, the equation can be said to define implicitly two different functions.

In many cases, it is difficult or impossible to solve an equation in $x$`x` and $y$`y` to give $y$`y` as a function of $x$`x` explicitly. Yet, we may be convinced that such a function exists and that it is differentiable.

We can differentiate term-by-term using the product rule and the fact that the derivative of $y$`y` (which we assume to be an implicitly defined differentiable function of $x$`x`) is just $\frac{\mathrm{d}y}{\mathrm{d}x}.$`d``y``d``x`.

The unit circle equation $x^2+y^2=1$`x`2+`y`2=1 defines two functions, one for the upper semicircle and one for the lower semicircle. These are both smooth curves. We can differentiate the equation implicitly.

Differentiating term-by-term, we have

$2x+2y.\frac{\mathrm{d}y}{\mathrm{d}x}=0$2`x`+2`y`.`d``y``d``x`=0. (We treated the $y^2$`y`2 term as a function of a function.)

On re-arranging, we find $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}$`d``y``d``x`=−`x``y`, provided $y\ne0.$`y`≠0. (The derivative is undefined at the end-points of the semicircles where the tangents are verticle.)

We could have differentiated the explicit form $y=\sqrt{1-x^2}$`y`=√1−`x`2. Thus, $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{\sqrt{1-x^2}}$`d``y``d``x`=−`x`√1−`x`2 but note that this is the same as $-\frac{x}{y}$−`x``y` when $y=\sqrt{1-x^2}$`y`=√1−`x`2 is substituted. A similar thing happens with the other explicit function, $y=-\sqrt{1-x^2}$`y`=−√1−`x`2.

Differentiate implicitly $x^2y+2xy^2=3$`x`2`y`+2`x``y`2=3.

By the product rule, we have$2xy+x^2\frac{\mathrm{d}y}{\mathrm{d}x}+2y^2+4xy\frac{\mathrm{d}y}{\mathrm{d}x}=0$2`x``y`+`x`2`d``y``d``x`+2`y`2+4`x``y``d``y``d``x`=0.

Therefore, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-2y(x+y)}{x(x+4y)}$`d``y``d``x`=−2`y`(`x`+`y`)`x`(`x`+4`y`).

In this example, it is possible to discover the explicit functions that satisfy the equation. We assume $y$`y` can be given as one or more functions of $x$`x`. Using the quadratic formula,

$x^2y+2xy^2$x2y+2xy2 |
$=$= | $3$3 |

$\therefore\ 2xy^2+x^2y-3$∴ 2xy2+x2y−3 |
$=$= | $0\ \ \text{This is a quadratic in y.}$0 This is a quadratic in y. |

$\therefore\ y$∴ y |
$=$= | $\frac{-x^2\pm\sqrt{x^4+24x}}{4x}$−x2±√x4+24x4x |

Thus, the equation is satisfied by two functions. The red and green curves in the diagram below represent the two functions.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems