y = -3/5x+7/5y=−35x+75 gives yy explicitly as a function of xx.
3x+5y=73x+5y=7 gives exactly the same relationship between xx and yy, but the function is implicit (hidden) in the equation. To make the function explicit, we solve for xx
In x^2+y^2=25x2+y2=25, yy is not a function of xx. However, there are two functions implicit in the equation. We can make the functions explicit by solving for yy.
y = +- sqrt(25-x^2)y=±√25−x2 is equivalent to the equation above and it has 2 functions that are not too difficult to make explicit:
y=sqrt(25-x^2)y=√25−x2 gives yy as a function of xx and
y=-sqrt(25-x^2)y=−√25−x2 gives yy as a different function of xx.
We can differentiate either the implicit or explicit presentations.
Differentiating implicitly (leaving the functions implicit) we get
2x+2y dy/dx = 02x+2ydydx=0 " " so " " dy/dx = -x/ydydx=−xy
The yy in the formula for the derivative is the price we pay for not making the function explicit. It replaces the explicit form of the function, whatever that may be.
For y=sqrt(25-x^2)y=√25−x2, we get dy/dx = - x/sqrt(25-x^2)dydx=−x√25−x2 (use the power and chain rule), and
for y= - sqrt(25-x^2)y=−√25−x2, we get dy/dx = x/sqrt(25-x^2)dydx=x√25−x2.
The equation y^5+4x^2y^2-3y+7x=28 y5+4x2y2−3y+7x=28 cannot be solved algebraically for yy, (or anyway, some 5th degree equations cannot be solved) but there are several functions of xx implicit in the equation. You can see them in the graph of the equation (shown below).
graph{y^5+4x^2y^2-3y+7x=28 [-7.14, 6.91, -4.66, 2.36]}
We can cut the graph into pieces, each of which is the graph of some function of xx on some domain.
Implicit differentiation allow us to find the derivative(s) of yy with respect to xx without making the function(s) explicit. Doing that, we can find the slope of the line tangent to the graph at the point (1,2)(1,2).
