How to prove that $\frac{d}{dx}\epsilon x^2=2\varepsilon x$ where $\varepsilon$ is just a constant

Question

How to prove that $\frac{d}{dx}\varepsilon x^2=2\varepsilon x$ where $\varepsilon$ is just a constant?

Thanks in advance for your immense help.

Answer 1

If $f$ is differentiable, and $c$ is constant, $$\frac{c f(x+h)-cf(x)}h=c\frac{f(x+h)-f(x)}h$$ Hence $(cf)'=cf'$ by letting $h\to 0$.

Answer 2

Hint: Use the product rule: $$\dfrac{\mathrm d}{\mathrm dx}\big(\epsilon\cdot x^2\big)=\left[\dfrac{\mathrm d}{\mathrm dx}\epsilon\right]x^2+\left[\dfrac{\mathrm d}{\mathrm dx}x^2\right]\epsilon.$$

Answer 3

We can prove this by using the product rule:

Let $f(x) = \epsilon, \quad g(x) = x^2$.

Then $$\dfrac{d}{dx}\Big(f(x)g(x)\Big) = f'(x)g(x) + f(x) g'(x)$$

and using the fact that $f'(x) = \dfrac d{dx}(\epsilon) = 0:\quad$ because the derivative of any constant is zero.

That said, it is imperative that you learn $$\frac{d}{dx}\left(ax^2\right) = 2ax$$ for any constant $a$.