Differential Calculus

Question

Find f'(x) from first principles given that f(x) is 

 

a) x^2√x

Answer 1

Given the function x²* sqrt(x) = f(x), we can simplify this using exponents to a single expression involving x...

Sqrt(x) = x^1/2

So, f(x) can be represented as x² * x^1/2 = x^5/2

Now using furst principles, we use Newton's method...

f'(x) = lim x->0 [ f(x + h) - f(x) ]/ h

f(x+h) = (x+h)^5/2 = (x+h)² * sqrt(x+h)

f(x+h) = (x² + 2xh + h²) * sqrt(x+h)

The sqrt function complicates the issue so we will attempt to rewrite Netwon's method to mitigate the effect if the sqrt function...

Multiplying by the number 1 doesnt change the function...

f'(x) = lim x->0 [ (f(x+h) - f(x)) * ( f(x+h) + f(x) )/( f(x+h) + f(x) )]/ h

f'(x) = lim x->0 [ (f(x+h)² - f(x)²) ]/ (h * (f(x+h) + f(x))

Looking at the numerator gives...

(x+h)⁵ - x⁵ = x⁵ + 5x⁴h + 10x³h² + 10x²h³ + 5xh⁴ + h⁵ - x⁵

= 5x⁴h + 10x³h² + 10x²h³ + 5xh⁴ + h⁵

Looking at the denominator gives...

f(x+h) + f(x) = (x+h)²sqrt(x+h) + x²sqrt(x)

Now dividing by h from the denominator into the numerator leaves..

f'(x) = lim x->0 [5x⁴ + 10x³h + 10x²h² + 5xh³ + h⁴]/(x+h)²sqrt(x+h) + x²sqrt(x)

f'(x) = 5x⁴ / 2x²sqrt(x) = 5/2 * x² / sqrt(x) = 5/2 * x^3/2

^