# Math Insight

### Derivatives of more general power functions

It's important to remember some of the other possibilities for the exponential notation $x^n$. For example $$x^{1\over 2}=\sqrt{x}$$ $$x^{-1}={1\over x}$$ $$x^{-{1\over 2}}={ 1 \over \sqrt{x }}$$ and so on. The good news is that the rule given just above for taking the derivative of powers of $x$ still is correct here, even for exponents which are negative or fractions or even real numbers: $${d\over dx}x^r=r\,x^{r-1}$$ Thus, in particular, $${d\over dx}\sqrt{x}= {d\over dx}x^{1\over 2}={1\over 2}x^{-{1\over 2}}$$ $${d\over dx}{1\over x}={d\over dx}x^{-1}=-1\cdot x^{-2}={-1\over x^2}$$

When combined with the sum rule and so on from above, we have the obvious possibilities: $${d\over dx}(3x^2-7\sqrt{x}+{5\over x^2}= {d\over dx}(3x^2-7x^{1\over 2}+5x^{-2})= 6x-{7\over 2}x^{-{1\over 2}}-10x^{-3}$$

The possibility of expressing square roots, cube roots, inverses, etc., in terms of exponents is a very important idea in algebra, and can't be overlooked.

#### Exercises

1. Find ${ d \over dx }(3x^7+5\sqrt{x}-11)$
2. Find ${ d \over dx }( {2\over x} + 5{\root 3 \of x} + 3)$
3. Find ${ d \over dx }( 7 - {5\over x^3} + 5x^7)$