The power rule
Rewrite each term as a constant multiplied by a power of x. Multiply the coefficient by the exponent, reduce the exponent by one and differentiate constants to zero
- Rewrite roots and reciprocals Write roots as fractional powers and reciprocals as negative powers before applying the rule \(\displaystyle \sqrt{x}=x^{1/2},\quad \frac{1}{x^{2}}=x^{-2}\)
- Multiply the coefficient and exponent For a term ax^n, the new coefficient is a multiplied by n. \(\displaystyle 5x^{3}\quad\Rightarrow\quad 5\times3x^{2}=15x^{2}\)
- Reduce the exponent by one After multiplying the coefficient, replace n by n-1 in the power \(\displaystyle x^{n}\quad\Rightarrow\quad nx^{n-1}\)
- Drop constant terms A constant has no change as x changes, so its derivative is zero \(\displaystyle \frac{d}{dx}\left(-7\right)=0\)
Worked example
Differentiate the function shown and find its derivative at the given input
- \(\displaystyle g(x)=4x^{1/2}+6x^{-2}-3x+8\)Rewrite the root and reciprocal as powers of x
- \(\displaystyle g'(x)=2x^{-1/2}-12x^{-3}-3\)Apply the power rule to each variable term and give the constant derivative as zero
- \(\displaystyle g'(1)=2(1)^{-1/2}-12(1)^{-3}-3\)Substitute the given input into the derivative, rather than the original function
- \(\displaystyle g'(1)=2-12-3=-13\)Simplify the three derivative terms to obtain -13
Common mistakes
Difficulty levels
Questions about this topic
Why are roots and reciprocals rewritten as powers?+
At inputs where the real power function is differentiable, roots and reciprocals can be written as powers and differentiated with the power rule. Check the original domain and any endpoint separately
What happens to a constant term when it is differentiated?+
It becomes zero because changing x does not change a constant
Why do I differentiate before substituting the input?+
The derivative is a different function that gives the gradient. Differentiate first, then substitute the input
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 5: Differentiation