Find the derivative of a power function Calculus

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = \dfrac{5}{2} and b = \dfrac{2}{5}. Therefore:

f'\left(x\right) = \dfrac{5}{2}.\dfrac{2}{5}x^{\frac{2}{5}-1}

f'\left(x\right) = x^{\frac{-3}{5}}

f'\left(x\right) = \dfrac{1}{x^{\frac{3}{5}}}

f'\left(x\right) = \dfrac{1}{x^{0.6}}

Note that the derivative's denominator is zero at x = 0, and therefore undefinined at x = 0.

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = -5 and b = 3. Therefore:

f'\left(x\right) = \left( 3 \right) \cdot \left(-5\right)x^{\left( 3 \right)-1}

f'\left(x\right) = -15x^{2}

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = \dfrac{1}{2} and b = \dfrac{-1}{2}. Therefore:

f'\left(x\right) = \left( \dfrac{-1}{2} \right) \cdot \dfrac{1}{2}x^{\left( \frac{-1}{2} \right)-1}

f'\left(x\right) = \dfrac{-1}{4}x^{\frac{-3}{2}}

f'\left(x\right) = -0.25x^{-1.5}

f'\left(x\right) = \dfrac{-0.25}{x^{1.5}}

Note that the derivative's denomiantor is zero at x = 0, and therefore undefinined at x = 0.

First of all notice that because \dfrac{1}{x} = x^{-1} the function is equivalent to :

f : x \longmapsto 9x^{-1}

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = 9 and b = -1

f'\left(x\right) = \left( -1 \right) \cdot 9x^{\left( -1 \right)-1}

f'\left(x\right) = -9x^{-2}

f'\left(x\right) = \dfrac{-9}{x^2}

Note that the derivative's denomiantor is zero at x = 0, and therefore undefinined at x = 0.

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = 3 and b = 3. Therefore:

f'\left(x\right) = \left( 3 \right) \cdot 3x^{\left( 3 \right)-1}

f'\left(x\right) = 9x^{2}

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = \dfrac{1}{4} and b = \dfrac{1}{2}. Therefore:

f'\left(x\right) = \left( \dfrac{1}{2} \right) \cdot \dfrac{1}{4}x^{\left( \frac{1}{2} \right)-1}

f'\left(x\right) = \dfrac{1}{8} x^{-\frac{1}{2}}
f'\left(x\right) = \dfrac{1}{8} x^{-0.5}

f'\left(x\right) = \dfrac{1}{8x^{0.5}}

Note that the derivative's denomiantor is zero at x = 0, and therefore undefinined at x = 0.

The derivative of a function of the form x\mapsto ax^b is:

x\mapsto \dfrac{d}{dx}ax^b = abx^{b-1}

Here, a = \left( \dfrac{2}{3} \cdot 2 \right) and b = \dfrac{3}{2}. Therefore:

f'\left(x\right) = \left( \dfrac{3}{2} \right) \cdot \left( \dfrac{2}{3} \cdot 2 \right)x^{\left( \frac{3}{2} \right)-1}

f'\left(x\right) = \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot x^{\frac{1}{2}}

f'\left(x\right) = 2x^{0.5}