Find the first derivative of each of the following functions

Find the first derivative of each of the following functions: a. f(x) = 2x5 + 4x4 – 6x2 + 1

b. f(x) = (x2 – 5x + 2)(2x2 – x + 4 ) --- Use the Product Rule

c. f(x) = (x2 – 1) / (x + 1) --- Use the Quotient Rule

d. f(x) = ((6x2 + 5) +3 )2 --- Use the Power Function Rule (Chain Rule)

e. f(x) = (8 –2x) (4-2x) x

f. f(X) = 40x + 1500 / x

g. f(x) = (x – 25) (130 – x)2

Solution

a.Given  f(x) = 2x⁵+ 4x⁴– 6x²+ 1

f\'(x) = d/dx( 2x⁵ + 4x⁴ – 6x² + 1)

= 10x⁴  + 16x³ - 12x

b. Given f(x) = (x² – 5x + 2)(2x² – x + 4 )

We know product rule D(uv) = uv\' + u\'v

f\'(x) = (x² – 5x + 2)*d/dx(2x² – x + 4 ) +d/dx (x² – 5x + 2)*(2x² – x + 4 )

= (x² – 5x + 2)(4x – 1 ) + (2x – 5)(2x² – x + 4 )

c. Given f(x) = (x² – 1) / (x + 1)

we know quotient rule D(u/v) = ( u\'v - uv\' )/v²

f\'(x) = [ d/dx(x² – 1) * (x + 1) - (x² – 1)* d/dx(x + 1) ] / (x + 1)²

= [ (2x)*(x + 1) - (x² – 1)(1) ]/(x + 1)²

   = [ x² + 2x - 1 ] / (x + 1)²

d. Given  f(x) = ((6x² + 5) +3 )²

we know power function rule D(xⁿ) = nx^n-1

f\'(x) = d/dx[ ((6x² + 5) +3 )² ]

= 2((6x² + 5) +3 ) * d/dx( (6x² + 5) +3 )

= 2((6x² + 5) +3 ) * (12x)

= 24x((6x² + 5) +3 )

e. Given   f(x) = (8 –2x) (4-2x) x

f\'(x) = d/dx[ (8 –2x) (4x-2x²) ]

= ( 8 - 2x)*( 4- 4x ) + (-2)*(4x-2x²)

= 32 - 40x + 8x² - 8x + 4x²

= 12x² - 48x + 32

f. Given  f(X) = 40x + 1500 / x

f\'(x) = d/dx [ 40x + 1500 / x ]

= 40 - 1500/x²

g. Given   f(x) = (x – 25) (130 – x)²

f\'(x) = d/dx[  (x – 25) (130 – x)² ]

= (x - 25)*d/dx[ (130 – x)² ] + d/dx[x-25]*(130 – x)²

= -2(x-25)(130-x) + (130 – x)²