If it's not what You are looking for type in the equation solver your own equation and let us solve it.
D( y )
y = 0
y = 0
y = 0
y in (-oo:0) U (0:+oo)
12*y^2-((y^-1)/2) = 0
12*y^2-1/2*y^-1 = 0
y^-1*(12*y^3-1/2) = 0
12*y^3 = 1/2 // : 12
y^3 = 1/24
y^3 = 1/24 // ^ 1/3
y = (1/24)^(1/3)
1/y = 0
1*y^-1 = 0 // : 1
y^-1 = 0
y należy do O
y = (1/24)^(1/3)
| 6x^2-4x=-10 | | 7x^2+2x-9= | | (y+3)(5y-4)= | | 5y+5-4+4y=12 | | X^4+4x+4=0 | | y^(-2)=25/4 | | x^2+2y^2=17 | | 42=-2(m+60)+m | | 75.36/18 | | 2p+5q=-49 | | 6x^3+9x^2=0 | | f(x)=4x^2+40x+3 | | 2x+12x=3x-5c | | 20t-5*t^2/2=30 | | 10x+7=-5x-7 | | 8x-x^2=6 | | (7^2)^3/(7^2*7^3) | | 33x+5x^2=0 | | cos6x/cos2x | | 2x+8=8x+9 | | 3x-5+15y+2x=2x+z | | y=4(-3)+11 | | x^2-80x=-200 | | -3x-9=x+1 | | 10y(2x)=2 | | 6x+3(4x+11)=-21 | | (3x-4)-(7x-9)=0 | | 3x+-5=6x+7 | | 16+2h=3h-15 | | (X^2-y^2-z^2)p+2xyq=z(x^2-y^2) | | x-1.8=4.2 | | Log[2](x^3+8x^2+11x-12)=3 |