NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Use implicit differentiation
Lesson

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

Example 1

Consider the function(s) that satisfy the equation $x^2y+xy^2=1$x2y+xy2=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$2xy+x2dydx+y2+2xydydx=0.

Hence, $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{y^2+2xy}{x^2+2xy}$dydx=y2+2xyx2+2xy

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 $\frac{\mathrm{d}y}{\mathrm{d}x}=0$dydx=0, we have $-\frac{y^2+2xy}{x^2+2xy}=0$y2+2xyx2+2xy=0 which is true if $y^2+2xy=0$y2+2xy=0. This equation is satified if $y=0$y=0 or if $y=-2x$y=2x.

On referring back to the original equation, we see that $y$y is never zero since no value of $x$x can make $x^2\times0+x\times0^2=1$x2×0+x×02=1 a true statement.

However, if we substitute $y=-2x$y=2x into the original equation, we get $-2x^3+4x^3=1$2x3+4x3=1. Hence, $2x^3=1$2x3=1 and so, $x=\frac{1}{\sqrt[3]{2}}$x=132 and then, $y=-\frac{2}{\sqrt[3]{2}}$y=232.

We have found a point where the gradient is zero, belonging to a function $y(x)$y(x) that satisfies the equation $x^2y+xy^2=1$x2y+xy2=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.

 

 

Example 2

Investigate the equation $x^2+xy+y^2=1.$x2+xy+y2=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$2x+y+xdydx+2ydydx=0.

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

At $(0,1),\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{2}$(0,1),dydx=12. So, the tangent at this point has equation $y=1-\frac{x}{2}$y=1x2.
At $(0,-1),\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{2}$(0,1),dydx=12. So, the tangent at this point has equation $y=-1-\frac{x}{2}$y=1x2
At $(1,0),\frac{\mathrm{d}y}{\mathrm{d}x}=-2$(1,0),dydx=2. The tangent here has equation $y=-2x+2$y=2x+2.
At $(-1,0),\frac{\mathrm{d}y}{\mathrm{d}x}=-2$(1,0),dydx=2. And, the tangent here has equation $y=-2x-2$y=2x2.

 

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

By substituting $y=-2x$y=2x into the original equation we have $x^2-2x^2+4x^2=1$x22x2+4x2=1. Hence, $3x^2=1$3x2=1 and so, $x=\pm\frac{1}{\sqrt{3}}$x=±13. The corresponding values of $y$y must be $\mp\frac{2}{\sqrt{3}}$23 and the horizontal tangent lines are therefore  $y=-\frac{2}{\sqrt{3}}$y=23 and $y=\frac{2}{\sqrt{3}}$y=23.

 

The graph of this relation is shown below with the tangent lines that we have found. The equation $x^2+xy+y^2=1$x2+xy+y2=1 defines an ellipse. Two functions are defined implicitly in the equation. They are shown in blue and red in the diagram.

 

 

 

Outcomes

M8-11

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

91578

Apply differentiation methods in solving problems

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