Hota to differentiate x^-3/4 using first principle?

Answer 1

First principles of differentiation deals with understanding if the function you are attempting to differentiate is continuous about the point of interest. In this case, since no point (in x) is specified, we will determine at which points of x the derivative is not well defined.

f(x) = x^(-3/4) needs to be checked for continuity (leading premise for differentiation)...

To test continuity we will determine if f(x) has an points in x which are not defined.

x = 0 is the only point where f(x) is ill-defined (i.e. ±∞)

lim (x-->0^-) f(x) = -∞

lim (x--> 0⁺) f(x) = +∞

Since there is no limit at x=0, the function is not continuous and therefore the derivative doesn't exist at x=0.

Otherwise,

lim (x-->a^-) f(x) = a(-3/4)

lim (x--> a⁺) f(x) = a(-3/4)

Since the limits are the same for real number a, we determine the f(x) is continuous except for a = 0

Using the definition of the derivative now...

lim (x-->a) ( f(x) - f(a) ) / ( x - a ) = f'(a)

Similarly we can recast into more convenient form as

lim (h-->0) ( f(x + h) - f(x) ) / h

Now we can substitute what we know...

f(x + h) = ⁴√(x+h)^-3

f(x) = ⁴√x^-3

f(x + h) - f(x) = ⁴√(x+h)^-3 - ⁴√x^-3

f(x + h) - f(x) * ( f(x+h) + f(x) ) / (f(x+h) + f(x) ) = f(x+h)² - f(x)² / ( f(x+h) + f(x) )

We don't want to forget about our new denominator factor...

f(x+h)² - f(x)² = √(x+h)^-3 - √x^-3

f(x+h)² - f(x)² * ( f(x+h)² + f(x)² ) / ( f(x+h)² + f(x)² ) = f(x+h)⁴ - f(x)⁴ / ( f(x+h)² + f(x)² )

We don't want to forget about our new denominator factor here either...

f(x+h)⁴ - f(x)⁴ = ( √(x+h)^-3 - √x^-3 ) * ( √(x+h)^-3 + √x^-3 ) = (x+h)^-3 - x^-3 = 1/(x+h)³ - 1/x³

Now we need to find common fraction --> x³/x³(x+h)³ - (x+h)³/(x+h)³x³ = ( -3x²h - 3xh² - h³ ) / (x³)(x+h)³

Putting this all together now...

lim (h-->0) ( f(x + h) - f(x) ) / h = [ ( -3x²h - 3xh² - h³ ) / (x³)(x+h)³ ]/h * 1/( [ ( f(x+h) + f(x) ) ]*[ ( f(x+h)² + f(x)² ) ] )

lim (h-->0) ( f(x + h) - f(x) ) / h = [ ( -3x² + 3xh - h² ) / (x³)(x+h)³ ] / ( [ ( f(x+h) + f(x) ) ]*[ ( f(x+h)² + f(x)² ) ] )

Now evaluate as h --> 0

1/( [ ( f(x+h) + f(x) ) ]*[ ( f(x+h)² + f(x)² ) ] ) = ( -3x²) / (x³)(x+h)³ ] / ( [ ( f(x) + f(x) ) ]*[ ( f(x)² + f(x)² ) ] ) = 1/(2 * f(x) * 2 * f(x)²)

1/( [ ( f(x+h) + f(x) ) ]*[ ( f(x+h)² + f(x)² ) ] ) = -3/x⁴ /(4 * f(x)³) = -3/4 * 1/(x⁴ * x^(-9/4)) = -3/4 * x^(9/4) / x⁴

1/( [ ( f(x+h) + f(x) ) ]*[ ( f(x+h)² + f(x)² ) ] ) = -3/4 * x^(-7/4)

Therefore f'(x) = lim (h-->0) ( f(x + h) - f(x) ) / h = -3/4 * x^(-7/4)