The derivative of power functions

In one of the previous exercises the "slope formulas", that is, the derivatives, were calculated for various power functions. We list some of them below, and always use powers:

function f→derivative f′x1→x0x2→2x1x3→3x2x1/2→12x−1/2x−1→−x−2\begin{array}{ccc} \text{function } f & \rightarrow & \text{derivative } f^\prime \\\hline x^1 & \rightarrow & x^0\\ x^2 & \rightarrow & 2x^1 \\ x^3 & \rightarrow & 3x^2 \\ x^{1/2} & \rightarrow & \frac{1}{2}x^{-1/2} \\ x^{-1} & \rightarrow & -x^{-2}\end{array}

Observe the emerging pattern. Apparently, for power functions it is very simple to deduct the formula for its derivative!!

Theorem 1

The derivative of the power function is obtained by decreasing the exponent by one and multiply by the original exponent:

Equation 1
f(x)=xn→′f′(x)=n⋅xn−1f(x)=x^n \overset{\prime}{\rightarrow} f^\prime(x)=n \cdot x^{n-1}

Power rule for differentiating power functions

This rule is actually correct for every exponent nn, but we will not give a prove. This rule is called the power rule of differential calculus (not to be mixed up with the power rules or power laws for multiplying powers ... but clearly you will need those soon enough).

Exercise 1

  1. Find the derivative of the following functions using the power rule:

    1. f(x)=x10f(x)=x^{10}
    2. g(x)=x3g(x)=\sqrt{x^3}
    3. h(x)=1x4h(x)=\frac{1}{x^4}
    4. l(x)=1x34l(x)=\frac{1}{\sqrt[4]{x^3}}

  2. Consider the function f(x)=x4/5f(x)=x^{4/5} and point AA on the graph of ff, where Ax=1A_x=1.

    1. Determine the slope of the tangent to the graph of ff at AA.

    2. Determine the equation of this tangent.

    3. Where does the tangent intersect the xx-axis?

    4. What is the angle of intersection between the tangent and the xx-axis?

Solution

  1. First, determine the exponent nn, then apply the rule:

    1. f(x)=x10f(x)=x^{10}, thus n=10n=10, thus f′(x)=10x9f^\prime(x)=10 x^9
    2. g(x)=x3=x3/2g(x)=\sqrt{x^3}=x^{3/2}, thus n=3/2n=3/2, thus g′(x)=32x0.5g^\prime(x)=\frac{3}{2} x^{0.5}
    3. h(x)=1x4=x−4h(x)=\frac{1}{x^4}=x^{-4}, thus n=−4n=-4, thus h′(x)=−4x−5=−4x5h^\prime(x)=-4 x^{-5}=-\frac{4}{x^5}
    4. l(x)=1x34=x−3/4l(x)=\frac{1}{\sqrt[4]{x^3}}=x^{-3/4}, thus n=−3/4n=-3/4, thus l′(x)=−34x−7/4=−34x74l^\prime(x)=-\frac{3}{4} x^{-7/4}=-\frac{3}{4\sqrt[4]{x^7}}

  2. We have the following:

    1. It is n=4/5n=4/5, so

      f′(x)=45x4/5−1=45x−1/5f^\prime(x)=\frac{4}{5} x^{4/5-1} = \frac{4}{5} x^{-1/5}

      The slope at x=1x=1 is therefore f′(1)=45⋅1−1/5=45‾f^\prime(1)=\frac{4}{5}\cdot 1^{-1/5}=\underline{\frac{4}{5}}.

    2. The equation of the tangent is t(x)=ax+bt(x)=ax+b, where a=45a=\frac{4}{5}. To find bb, note that we have

      t(1)=f(1)45⋅1+b=14/5b=15\begin{array}{rll} t(1) & = & f(1) \\ \frac{4}{5}\cdot 1+b & = &1^{4/5}\\ b &= & \frac{1}{5} \end{array}

      Thus, t(x)=45x+15‾t(x)=\underline{\frac{4}{5} x+\frac{1}{5}}.

    3. Find xx with t(x)=45x+15=0t(x)=\frac{4}{5} x+\frac{1}{5}=0. It follows x=−14‾x=\underline{-\frac{1}{4}}.

    4. The slope of the straight line tt is a=δyδx45a=\frac{\delta y}{\delta x}\frac{4}{5}, where δx\delta x and δy\delta y are the side lengths of the slope triangle. If α\alpha is the angle between the xx-axis and the tangent, we have

      tan⁡(α)=OA=ΔyΔx=45\tan(\alpha)=\frac{O}{A}=\frac{\Delta y}{\Delta x}=\frac{4}{5}

      and thus

      α=arctan⁡(45)=38.66∘‾\alpha=\arctan\left(\frac{4}{5}\right)= \underline{38.66^\circ}