derivative of (3x-2)^3(8x^2+7)^3
1) Let's differentiate: ddx((3x -2)3·(8x2 + 7)3)
For u1 = (3x -2)3 and v2 = (8x2 + 7)3, use the differentiation rule: ddx(u1·v2) = ddx(u1)·v2 + u1·ddx(v2)
2) Let's perform the differentiation u1: ddx((3x -2)3)
Apply the differentiation rule: ddx(fn) = n·ddx(f)·fn -1 with f = 3x -2
ⓘ The derivative of x^n is n*x^(n-1): ddx((3x -2)3) = 3·ddx(3x -2)·(3x -2)2
3·ddx(3x -2)·(3x -2)2
ⓘ The derivative of a sum is the sum of the derivatives: ddx(3x -2) = ddx(3x) + 0
3·(ddx(3x) + 0)·(3x -2)2
ⓘ Adding 0 changes nothing: ddx(3x) + 0 = ddx(3x)■A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(3x) = 3
3·3·(3x -2)2
ⅰ 3·3 = 9
9·(3x -2)2
3) Let's differentiate v2: ddx((8x2 + 7)3)
We differentiate using: ddx(fn) = n·ddx(f)·fn -1 (where f = 8x2 + 7)
ⓘ The derivative of x^n is n*x^(n-1): ddx((8x2 + 7)3) = 3·ddx(8x2 + 7)·(8x2 + 7)2
3·ddx(8x2 + 7)·(8x2 + 7)2
ⓘ The derivative of a sum is the sum of the derivatives: ddx(8x2 + 7) = ddx(8x2) + 0
3·(ddx(8x2) + 0)·(8x2 + 7)2
ⓘ Adding 0 changes nothing: ddx(8x2) + 0 = ddx(8x2)■A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(8x2) = 8·ddx(x2)■The derivative of x^n is n*x^(n-1): ddx(x2) = 2x
3·8·2x·(8x2 + 7)2
ⅰ 3·8 = 24
24·2x·(8x2 + 7)2
ⅰ 24·2 = (20 + 4)·2 = 20·2 + 4·2 = 40 + 8 = 48
48x·(8x2 + 7)2
4) Then you can resume the differentiation:
ⅰ ddx((3x -2)3·(8x2 + 7)3) = 9·(3x -2)2·(8x2 + 7)3 + (3x -2)3·48x·(8x2 + 7)2
9·(3x -2)2·(8x2 + 7)3 + (3x -2)3·48x·(8x2 + 7)2
48x·(3x -2)3·(8x2 + 7)2 + 9·(3x -2)2·(8x2 + 7)3
You can write the expression in factored form: (3x -2)2·((8x2 + 7))2·(48x·(3x -2) + 9·(8x2 + 7))
Let's factorize in ℝ: (48x·(3x -2) + 9·(8x2 + 7))
We factor out 3, the greatest common divisor of the coefficients: 48x·(3x -2) + 9·(8x2 + 7) = 3·(16x·(3x -2) + 3·(8x2 + 7))
Factor found: 16x·(3x -2) + 3·(8x2 + 7)
▶Derivative:   3·(3x -2)2·(8x2 + 7)2·(16x·(3x -2) + 3·(8x2 + 7))