Differentiation

Lesson

Implicit differentiation can be used to discover information about the gradients of functions defined implicitly.

Consider the function(s) that satisfy the equation $x^2y+xy^2=1$`x`2`y`+`x``y`2=1. Taking $y$`y` to be a function of $x$`x`, we can differentiate with respect to $x$`x` to obtain

$2xy+x^2\frac{\mathrm{d}y}{\mathrm{d}x}+y^2+2xy\frac{\mathrm{d}y}{\mathrm{d}x}=0$2`x``y`+`x`2`d``y``d``x`+`y`2+2`x``y``d``y``d``x`=0.

Hence, $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{y^2+2xy}{x^2+2xy}$`d``y``d``x`=−`y`2+2`x``y``x`2+2`x``y`

We may wish to know whether the gradient of the function $y$`y` is ever zero, as would happen at a maximum or minimum.

Putting $\frac{\mathrm{d}y}{\mathrm{d}x}=0$`d``y``d``x`=0, we have $-\frac{y^2+2xy}{x^2+2xy}=0$−`y`2+2`x``y``x`2+2`x``y`=0 which is true if $y^2+2xy=0$`y`2+2`x``y`=0. This equation is satified if $y=0$`y`=0 or if $y=-2x$`y`=−2`x`.

On referring back to the original equation, we see that $y$`y` is never zero since no value of $x$`x` can make $x^2\times0+x\times0^2=1$`x`2×0+`x`×02=1 a true statement.

However, if we substitute $y=-2x$`y`=−2`x` into the original equation, we get $-2x^3+4x^3=1$−2`x`3+4`x`3=1. Hence, $2x^3=1$2`x`3=1 and so, $x=\frac{1}{\sqrt[3]{2}}$`x`=1^{3}√2 and then, $y=-\frac{2}{\sqrt[3]{2}}$`y`=−2^{3}√2.

We have found a point where the gradient is zero, belonging to a function $y(x)$`y`(`x`) that satisfies the equation $x^2y+xy^2=1$`x`2`y`+`x``y`2=1. Further investigation would be needed to determine whether this point is a local maximum or local minimum or neither.

The graph of the functions defined implicitly by the given equation is shown below with the horizontal tangent shown in pink.

Investigate the equation $x^2+xy+y^2=1.$`x`2+`x``y`+`y`2=1.

By inspection, when $x=0$`x`=0, $y=\pm1$`y`=±1 and when $y=0$`y`=0, $x=\pm1$`x`=±1. So, the graph of this relation contains the points $(0,1),(0,-1),(1,0)$(0,1),(0,−1),(1,0) and $(-1,0).$(−1,0).

We could find the gradients at each of these points. Implicit differentiation with respect to $x$`x` gives $2x+y+x\frac{\mathrm{d}y}{\mathrm{d}x}+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0$2`x`+`y`+`x``d``y``d``x`+2`y``d``y``d``x`=0.

Thus, $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x+y}{x+2y}$`d``y``d``x`=−2`x`+`y``x`+2`y`.

At $(0,1),\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{2}$(0,1),`d``y``d``x`=−12. So, the tangent at this point has equation $y=1-\frac{x}{2}$`y`=1−`x`2.

At $(0,-1),\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{2}$(0,−1),`d``y``d``x`=−12. So, the tangent at this point has equation $y=-1-\frac{x}{2}$`y`=−1−`x`2

At $(1,0),\frac{\mathrm{d}y}{\mathrm{d}x}=-2$(1,0),`d``y``d``x`=−2. The tangent here has equation $y=-2x+2$`y`=−2`x`+2.

At $(-1,0),\frac{\mathrm{d}y}{\mathrm{d}x}=-2$(−1,0),`d``y``d``x`=−2. And, the tangent here has equation $y=-2x-2$`y`=−2`x`−2.

We could look for any points at which the gradient is zero by putting $\frac{\mathrm{d}y}{\mathrm{d}x}=0$`d``y``d``x`=0. Thus, $-\frac{2x+y}{x+2y}=0$−2`x`+`y``x`+2`y`=0 and this will be true if $2x+y=0$2`x`+`y`=0. That is, if $y=-2x$`y`=−2`x`.

By substituting $y=-2x$`y`=−2`x` into the original equation we have $x^2-2x^2+4x^2=1$`x`2−2`x`2+4`x`2=1. Hence, $3x^2=1$3`x`2=1 and so, $x=\pm\frac{1}{\sqrt{3}}$`x`=±1√3. The corresponding values of $y$`y` must be $\mp\frac{2}{\sqrt{3}}$∓2√3 and the horizontal tangent lines are therefore $y=-\frac{2}{\sqrt{3}}$`y`=−2√3 and $y=\frac{2}{\sqrt{3}}$`y`=2√3.

The graph of this relation is shown below with the tangent lines that we have found. The equation $x^2+xy+y^2=1$`x`2+`x``y`+`y`2=1 defines an ellipse. Two functions are defined implicitly in the equation. They are shown in blue and red in the diagram.

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