# How do I use the power rule with $4x^2 - 2$?

Every website I look at only shows the power use with a single term example or polynomial. Could somebody please give me a step by step breakdown of using the power rule to find $f'(x)$ where $f(x) = 4x^2 - 2$?

• The power rule is linear, so you can use it term by term, and then you can combine the results.
– Paul
Oct 9 '15 at 16:15
• Exactly, how does the power rule affect constants? I know that 4x^2 goes to 8x but how does -2 work? Oct 9 '15 at 16:16
• " single term example or polynomial": your case is a polynomial, why can't you derive it ?
– user65203
Oct 9 '15 at 16:20
• @danishanish: The derivative of any constant term is zero. This makes sense intuitively, since the derivative represents a rate of change, and a constant doesn't change. Oct 9 '15 at 17:03

First, the derivative is a linear operator. This means that $$(f(x) + g(x))' = f'(x) + g'(x)$$ and $$(cf)'(x) = cf'(x)$$ So, in the case that you give, $h(x) = 4x^2 - 2$, you can look at this as $$h(x) = f(x) + g(x) \quad \text{where} \quad f(x) = 4x^2 \quad \text{and} \quad g(x) = -2$$ Then, you have \begin{align*} h'(x) &= (f(x) + g(x))' \\ &= f'(x) + g'(x) \\ &= (4x^2)' + (-2)' \\ &= 4(x^2)' + 0 \\ &= 4 (2x) \\ &= 8x. \end{align*}

Of course, once you practice taking the derivatives of polynomials, this will all seem like a single step. We could even add in a few more steps to the above to pull out the constants, etc. Hopefully this is clear already though.

• Gotcha. Thanks! Oct 9 '15 at 16:19
• "this will all seem like a single step." This is very important. As long as you need more than one second to derive a quadratic polynomial, you know you need to practice a lot more!
– 5xum
Oct 9 '15 at 16:29
• @5xum coming back a while later to respond, yes - deriving any quadratic polynomial is extremely simple and fast now - thanks for the help :) May 16 '16 at 5:05

There are already a ton of good answers to this, but I can't help myself:

Just for fun, since you wrote product rule in the title and because everyone else showed you the power rule, I'll show you how to use the product rule for this. ;)

Remember that the product rule is that $$(f(x)\cdot g(x))' = f'(x)\cdot g(x) + f(x)\cdot g'(x)$$

In your case we have $$h(x) = 4x^2-2$$ Using what we know about the difference of squares, we can factor this as \begin{align}h(x) &= 4x^2 -2 \\ &= (2x-\sqrt{2})(2x+\sqrt{2})\end{align}

Now we can use the product rule. Just let $\require{enclose}f(x) = 2x-\sqrt{2}$ and $g(x) = 2x+\sqrt{2}$. Then \begin{align}h'(x) &= \left(2x-\sqrt{2}\right)'\left(2x+\sqrt{2}\right) + \left(2x-\sqrt{2}\right)\left(2x+\sqrt{2}\right)' \\ &= \left[\left(2x\right)'-\left(\sqrt{2}\right)'\right]\left(2x+\sqrt{2}\right) + \left(2x-\sqrt{2}\right)\left[\left(2x\right)'+\left(\sqrt{2}\right)'\right] \\ &= \left[2-0\right]\left(2x+\sqrt{2}\right) + \left(2x-\sqrt{2}\right)\left[2+0\right] \\ &= 4x+2\sqrt{2} + 4x-2\sqrt{2} \\ &= \enclose{box}{8x}\end{align}

$(4x^2-2)'=(4x^2)'-(2)'=4(x^2)'-0=4\cdot 2x=8x$.

Remember that the derivative of a constant is always $0$.

$$\begin{split} \frac{df(x)}{dx} = \frac{d}{dx} \left[ 4x^2 - 2\right] = \frac{d\left[ 4x^2 \right]}{dx} - \frac{d [2]}{dx} = 4 \frac{d\left[ x^2 \right]}{dx} - 0 = 4 \cdot 2x = 8x \end{split}$$

Hint: Let $g(x)=2x$. Then $f(x)=4x^2=\left(g(x)\right)^2$ Therefore $f'(x)=2\cdot g(x)\cdot g'(x)$

There's a separate rule (the sum rule) for taking the derivative of any sum of terms:

(f(x) + g(x))' = f'(x) + g'(x)

Use this to split the sum into its terms and take the derivative of each term.