derivative of a/(bx^2)
⚠ bx is read as the product b*x: for the variable bx, write bx_
The expression abx2 contains several variables: the variable x is chosen for differentiation
1) Let's perform the differentiation: ddx(abx2)
For p1 = a and q2 = bx2, use the differentiation rule: ddx(p1q2) = ddx(p1)·q2 -p1·ddx(q2)q22
2) Let's perform the differentiation p1: ddx(a)
ⓘ The derivative of a constant is zero: ddx(a) = 0
0
3) Let's differentiate q2: ddx(bx2)
ⓘ For differentiation with respect to x, b is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(bx2) = b·ddx(x2)■The derivative of x^n is n*x^(n-1): ddx(x2) = 2x
b·2x
4) Let's evaluate the resulting derivative (p1q2)': 0·bx2 -a·b·2x(bx2)2
ⓘ A product with a zero factor is zero: 0·bx2 = 0
0 -ab·2x(bx2)2
ⓘ Adding 0 changes nothing: 0 -ab·2x = -ab·2x
-ab·2x(bx2)2
ⓘ The power of a product is the product of the powers: (bx2)2 = b2·(x2)2
-ab·2xb2·(x2)2
We simplify the fraction: bb2 by the common factor: b to get the new fraction: 1b
2·(-ax)b·(x2)2
ⓘ The minus sign is placed in front of the product: 2·(-a) = -2a■A power of a power multiplies the exponents, (a^n)^m = a^(n*m): (x2)2 = x2·2
-2axbx2·2
ⅰ 2·2 = 4
-2axbx4
We divide the numerator and denominator of xx4 by the common factor: x, which gives the new fraction: 1x3
-2abx3
-2abx3
5) Then you can continue the differentiation:
ⅰ ddx(abx2) = -2abx3
▶Derivative:   -2abx3