derivative of ((x^2-5)^3)/(x^2+2)
1) Let's differentiate: ddx((x2 -5)3x2 + 2)
For p1 = (x2 -5)3 and q2 = x2 + 2, use the differentiation rule: ddx(p1q2) = ddx(p1)·q2 -p1·ddx(q2)q22
2) Let's perform the differentiation p1: ddx((x2 -5)3)
We differentiate using: ddx(fn) = n·ddx(f)·fn -1 (where f = x2 -5)
ⓘ The derivative of x^n is n*x^(n-1): ddx((x2 -5)3) = 3·ddx(x2 -5)·(x2 -5)2
3·ddx(x2 -5)·(x2 -5)2
ⓘ The derivative of a sum is the sum of the derivatives: ddx(x2 -5) = ddx(x2) + 0
3·(ddx(x2) + 0)·(x2 -5)2
ⓘ Adding 0 changes nothing: ddx(x2) + 0 = ddx(x2)■The derivative of x^n is n*x^(n-1): ddx(x2) = 2x
3·2x·(x2 -5)2
ⅰ 3·2 = 6
6x·(x2 -5)2
3) Let's differentiate q2: ddx(x2 + 2)
ⓘ The derivative of a sum is the sum of the derivatives: ddx(x2 + 2) = ddx(x2) + 0
ddx(x2) + 0
ⓘ Adding 0 changes nothing: ddx(x2) + 0 = ddx(x2)■The derivative of x^n is n*x^(n-1): ddx(x2) = 2x
2x
4) We evaluate the resulting derivative (p1q2)': 6x·(x2 -5)2·(x2 + 2) -(x2 -5)3·2x(x2 + 2)2
ⓘ What the terms have in common is factored out, so that the fraction can be simplified: 6x·(x2 -5)2·(x2 + 2) -(x2 -5)3·2x(x2 + 2)2 = 2x·(3·(x2 + 2)·(x2 -5)2 -(x2 -5)3)(x2 + 2)2
2x·(3·(x2 + 2)·(x2 -5)2 -(x2 -5)3)(x2 + 2)2
ⓘ What the terms have in common is factored out, so that the fraction can be simplified: 2x·(3·(x2 + 2)·(x2 -5)2 -(x2 -5)3)(x2 + 2)2 = 2x·(-(x2 -5))2·(3·(x2 + 2) -(x2 -5))(x2 + 2)2
2x·(-(x2 -5))2·(3·(x2 + 2) -(x2 -5))(x2 + 2)2
ⓘ A minus sign before a parenthesis changes the sign of each of its terms: -(x2 -5) = -x2 + 5
2x·(-x2 + 5)2·(3·(x2 + 2) -x2 + 5)(x2 + 2)2
ⓘ Expand by multiplying each term inside the parentheses: 3·(x2 + 2) = 3x2 + 6
2x·(-x2 + 5)2·(3x2 + 6 -x2 + 5)(x2 + 2)2
ⓘ Parentheses are computed first: 6 + 5 = 11■Like terms are grouped by adding their coefficients: 3x2 -x2 = 2x2
2x·(-x2 + 5)2·(2x2 + 11)(x2 + 2)2
5) Then you can resume the differentiation:
ⅰ ddx((x2 -5)3x2 + 2) = 2x·(-x2 + 5)2·(2x2 + 11)(x2 + 2)2
2x·(-x2 + 5)2·(2x2 + 11)(x2 + 2)2
▶Derivative:   2x·(2x2 + 11)·(5 -x2)2(x2 + 2)2