derivative of (1-25x^2)^2arcsin(5x)
1) Let's perform the differentiation: ddx((1 -25x2)2·arcsin(5x))
For u1 = (1 -25x2)2 and v2 = arcsin(5x), use the differentiation rule: ddx(u1·v2) = ddx(u1)·v2 + u1·ddx(v2)
2) Let's differentiate u1: ddx((1 -25x2)2)
Apply the differentiation rule: ddx(fn) = n·ddx(f)·fn -1 with f = 1 -25x2
ⓘ The derivative of x^n is n*x^(n-1): ddx((1 -25x2)2) = 2·ddx(1 -25x2)·(1 -25x2)
2·ddx(1 -25x2)·(1 -25x2)
ⓘ Expand by multiplying each term inside the parentheses: 2·ddx(1 -25x2)·(1 -25x2) = 2·ddx(1 -25x2) -50·ddx(1 -25x2)·x2■The derivative of a sum is the sum of the derivatives: ddx(1 -25x2) = 0 + ddx(-25x2)
2·(0 + ddx(-25x2)) -50·(0 + ddx(-25x2))·x2
ⓘ Adding 0 changes nothing: 0 + ddx(-25x2) = ddx(-25x2)■A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(-25x2) = -25·ddx(x2)■The derivative of x^n is n*x^(n-1): ddx(x2) = 2x
2·(-25·2x) -50·(-25·2x)·x2
ⓘ Multiplications are done before additions and subtractions: 25·2 = (20 + 5)·2 = 20·2 + 5·2 = 40 + 10 = 50■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: -50·(-25) = 50·25 = 50·(20 + 5) = 50·20 + 50·5 = 1000 + 250 = 1250■Multiplying powers with the same base adds the exponents: xx2 = x3
-50·2x + 1250·2x3
ⓘ Multiplications are done before additions and subtractions: 50·2 = 100, 1250·2 = (1000 + 200 + 50)·2 = 1000·2 + 200·2 + 50·2 = 2000 + 400 + 100 = 2500
-100x + 2500x3
3) Let's differentiate v2: ddx(arcsin(5x))
Apply the differentiation rule: ddx(arcsin(f)) = ddx(f)·11 -f2 with f = 5x
ⓘ The derivative of arcsin(x) is 1/sqrt(1 - x^2): ddx(arcsin(5x)) = ddx(5x)1 -(5x)2
ddx(5x)1 -(5x)2
ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(5x) = 5■The power of a product is the product of the powers: (5x)2 = 52·x2
51 -52·x2
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: 52 = 5·5 = 25
51 -25x2
4) We evaluate the resulting derivative (u1·v2)': (-100x + 2500x3)·arcsin(5x) + (1 -25x2)2·51 -25x2
ⓘ Multiplying a fraction by a term multiplies its numerator: (1 -25x2)2·51 -25x2 = (1 -25x2)2·51 -25x2■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor: (1 -25x2)2·51 -25x2 = 5·(1 -25x2)2 -12
(-100x + 2500x3)·arcsin(5x) + 5·(1 -25x2)2 -12
ⓘ To add fractions, put them over the same denominator, then add the numerators: 2 -12 = 2·2 -12
(-100x + 2500x3)·arcsin(5x) + 5·(1 -25x2)2·2 -12
ⓘ Multiplications are done before additions and subtractions: 2·2 = 4
(-100x + 2500x3)·arcsin(5x) + 5·(1 -25x2)4 -12
ⓘ The numerator and the denominator are computed before the division: 4 -1 = 3
(-100x + 2500x3)·arcsin(5x) + 5·(1 -25x2)32
ⓘ Expand by multiplying each term inside the parentheses: (-100x + 2500x3)·arcsin(5x) = -100x·arcsin(5x) + 2500x3·arcsin(5x)
-100x·arcsin(5x) + 2500x3·arcsin(5x) + 5·(1 -25x2)32
5) Then you can resume the derivation:
ⅰ ddx((1 -25x2)2·arcsin(5x)) = -100x·arcsin(5x) + 2500x3·arcsin(5x) + 5·(1 -25x2)32
-100x·arcsin(5x) + 2500x3·arcsin(5x) + 5·(1 -25x2)32
2500x3·arcsin(5x) -100x·arcsin(5x) + 5·(1 -25x2)32
We factor out 5, the greatest common divisor of the coefficients: 2500x3·arcsin(5x) -100x·arcsin(5x) + 5·(1 -25x2)32 = 5·(500x3·arcsin(5x) -20x·arcsin(5x) + (1 -25x2)32)
Factor identified: 500x3·arcsin(5x) -20x·arcsin(5x) + (1 -25x2)32
▶Derivative:   5·(500x3·arcsin(5x) -20x·arcsin(5x) + (1 -25x2)32)