derivative of y=(5x-2)/(sqrt(x))
We save the definition: y = 5x -2x
1) Let's perform the differentiation: ddx(5x -2x)
We differentiate p1 = 5x -2 and q2 = x using: ddx(p1q2) = ddx(p1)·q2 -p1·ddx(q2)q22
2) Let's differentiate p1: ddx(5x -2)
ⓘ The derivative of a sum is the sum of the derivatives: ddx(5x -2) = ddx(5x) + 0
ddx(5x) + 0
ⓘ Adding 0 changes nothing: ddx(5x) + 0 = ddx(5x)■A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(5x) = 5
5
3) Let's differentiate q2: ddx(x)
ⓘ The derivative of sqrt(x) is 1/(2*sqrt(x)): ddx(x) = 12x
12x
4) Let's evaluate the resulting derivative (p1q2)': 5x -(5x -2)·12x(x)2
ⓘ Multiplying a fraction by a term multiplies its numerator: (5x -2)·12x = (5x -2)·12x = 5x -22x■To add fractions, put them over the same denominator, then add the numerators: 5x -5x -22x = 5x·2x -(5x -2)2x■A power of a power multiplies the exponents, (a^n)^m = a^(n*m): (x)2 = x12·2
5x·2x -(5x -2)2xx12·2
ⓘ A minus sign before a parenthesis changes the sign of each of its terms: -(5x -2) = -5x + 2■Multiplications are done before additions and subtractions: 5·2 = 10■Multiplying powers with the same base adds the exponents: x·x = x■Multiplying a fraction by a number multiplies its numerator: 12·2 = 1·22 = 22■We simplify the fraction by dividing the numerator and the denominator by 2, their greatest common divisor: 22 = 1
10x -5x + 22xx1
ⓘ Like terms are grouped by adding their coefficients: 10x -5x = 5x
5x + 22xx
ⓘ Dividing a fraction by x is multiplying its denominator by x: 5x + 22xx = 5x + 22x·x
5x + 22x·x
ⓘ Multiplying powers with the same base adds the exponents: xx = x12 + 1
5x + 22x12 + 1
ⓘ To add fractions, put them over the same denominator, then add the numerators: 12 + 1 = 1 + 22
5x + 22x1 + 22
ⓘ The numerator and the denominator are computed before the division: 1 + 2 = 3
5x + 22x32
5) Then you can continue the derivation:
ⅰ ddx(5x -2x) = 5x + 22x32
▶Derivative:   5x + 22x32