Let f(x)=(4x-5)^(1/2). Find formula for f^(n)x

Question

Find a formula for f^n

Answer 1

Differentiate, repeatedly, to recognize the pattern that evolves.

 

f(x)=(4x-5)^½

 

f'(x) = 2(4x-5)^-½ by application of the power rule and the chain rule.

 

f"(x) = -4(4x-5)^-3/2

 

f'''(x) = 24(4x-5)^-5/2

 

f⁽⁴⁾(x) = -240(4x-5)^-7/2

 

The pattern is easy as far as the binomial and its power; analyze the coefficient by looking at prime factorizations, etc--I leave this "little" task to you.

Answer 2

Need to find a pattern and then perhaps prove it by induction....

 

f'(x) = (1/2)(4x-5)^(-1/2)(4) = 2(4x-5)^(-1/2)

 

f''(x) =  -(4x-5)^(-3/2)(4) = (-4)(4x-5)^(-3/2)

 

f'''(x) = (6)(4x-5)^(-5/2)(4) = (24)(4x-5)^(-5/2)

 

f4(x) = (-60)(4x-5)^(-7/2)(4) = (-240)(4x-5)^(-7/2)

 

f5(x) = (840)(4x-5)^(-9/2)(4) = (3360)(4x-5)^(-9/2)

 

 

The exponents of (4x-5) are (-2n+1)/2

 

The coefficients increase by a factor of 4 more than the previous and alternate signs.

Specifically,

   4/2 = 2; 

  24/4 = 6;   <---- 4 more than the previous factor

 240/24 = 10 <--- 4 more than the previous factor

 3360/240 = 14 <--- 4 more than the previous factor

 

So for the 6th derivative, we expect the coefficient to be -1*3360 x 18 = -60480

 

Differentiating, f6(x) = 3360/2*-9 (4x-5)^(-11/2) (4) = 3360*18 (4x-5)^(-11/2) = -60480(4x-5)^(-11/2)