Derivatives of 1/x and √x

$(x^n)'=nx^{n-1}$ also holds when the exponent is negative or a fraction. For $\frac{1}{x}=x^{-1}$ you get $-x^{-2}=-\frac{1}{x^2}$, and for $\sqrt{x}=x^{\frac{1}{2}}$ you get $\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

When the inside is $ax+b$, the chain rule multiplies by the derivative of the inside, $a$.

For example, $\left(\sqrt{4x+5}\right)'=\frac{4}{2\sqrt{4x+5}}=\frac{2}{\sqrt{4x+5}}$. Reduce the fraction when you can.

Examples

1. 01Differentiate$\displaystyle \frac{d}{dx} \frac{1}{x}$AnswerHide

   Answer$\displaystyle -\frac{1}{x^2}$
2. 02Differentiate$\displaystyle \frac{d}{dx} \sqrt{5x + 1}$AnswerHide

   Answer$\displaystyle \frac{5}{2\sqrt{5x + 1}}$
3. 03Differentiate$\displaystyle \frac{d}{dx} \sqrt{4x + 5}$AnswerHide

   Answer$\displaystyle \frac{2}{\sqrt{4x + 5}}$
4. 04Differentiate$\displaystyle \frac{d}{dx} \sqrt{2x + 3}$AnswerHide

   Answer$\displaystyle \frac{1}{\sqrt{2x + 3}}$
5. 05Differentiate$\displaystyle \frac{d}{dx} \sqrt{x + 6}$AnswerHide

   Answer$\displaystyle \frac{1}{2\sqrt{x + 6}}$
6. 06Differentiate$\displaystyle \frac{d}{dx} \sqrt{2x + 6}$AnswerHide

   Answer$\displaystyle \frac{1}{\sqrt{2x + 6}}$

FAQ

Q1I got $\frac{1}{2\sqrt{5x+1}}$ for the derivative of $\sqrt{5x+1}$. What went wrong?AnswerHide

A

You forgot to multiply by $5$, the derivative of the inside $5x+1$. The answer is $\frac{5}{2\sqrt{5x+1}}$.

Q2Why is the derivative of $\frac{1}{x}$ negative?AnswerHide

A

Since $\frac{1}{x}=x^{-1}$, the exponent $-1$ comes down in front, giving $-x^{-2}$. On the graph, the curve goes down as you move right on both $x>0$ and $x<0$ (negative slope).