thharrimw said:when i took the derivative of X^2*Y^2-2x=6 i didn't get the same thing that i got when i took the derivative of y^2=(6+2x)/x^2 and all i did was rewrite the origional equation. why did rewriting the equation change my answer?
Okay, [itex]x^2y^2- 2x= 3 has derivative [itex]2xy^2+ 2x^2y y'- 2= 0 so [itex]2x^2y y'= 2- 2xy^2[/itex] [itex]y'= (2-2xy^2)/(2x^2y)[/itex], the "2"s cancel and y'= (1- xy^2)/(x^2y). You seem to have lost the "2" in the numerator.thharrimw said:not rewritting i gou
(x^2)(Y^2)-2x=3
(x^2)(2y)(y')+2x(y^2)-2=0
(x^2)(2y)(y')=2-(2x)(Y^2)
y'=(1-x(y^2))/((x^2)2y))
No, solving for y2 gives [itex]x^2 y^2=6+ 2x[/itex] so [itex]y^2= (6+ 2x)/x^2[/itex] The derivative of [itex]y^2[/itex], with respect to x, is 2y y' and the derivative of [itex](6+2x)/x^2[/itex] using the quotient rule is [itex][(2)x^2- (6+2x)(2x)]/(x^4)= [-2x^2- 12x]/(x^4)= -(2x^2+ 12x)/(x^4)= -2(x+ 6)/(x^3)[/itex]. [itex]2yy'= -2(x+6)/x^3[/itex] gives [itex]y'= -(x+1)/yx^3[/itex].thharrimw said:when i rewrote it i got y^2=(3+2x)/(x^2) and my derivitive was (x-3)/(y(x^3))
thharrimw said:when i rewrote it i got y^2=(3+2x)/(x^2)
and my derivitive was (x-3)/(y(x^3))
A "weird implicit derivative" refers to the derivative of a function that cannot be expressed explicitly in terms of its variables. This typically occurs when the function is defined implicitly, meaning that one variable is defined in terms of another.
The "weird implicit derivative" differs from a regular derivative in that it cannot be found by simply applying the power rule or chain rule. Instead, implicit differentiation must be used to find the derivative of the function.
The "weird implicit derivative" is important because it allows us to find the instantaneous rate of change of a function even when it cannot be expressed explicitly. This is useful in many applications, such as in physics and engineering.
Some common techniques for finding the "weird implicit derivative" include using the chain rule, product rule, and quotient rule. Another approach is to solve the implicit equation for one variable and then take the derivative with respect to the other variable.
Yes, the "weird implicit derivative" can be negative. This indicates that the function is decreasing in that particular region. However, it is important to note that the sign of the "weird implicit derivative" does not necessarily correspond to the sign of the regular derivative.