Question on differentiation/finding the derivative of an equation (in calculus)

Question

The equation of a parabola is given as $y= 3x^2 - 12x$

• Find $\frac{\mathrm{d}y}{\mathrm{d}x}$

$$\begin{align}y=& 3x^2 - 12x\\ =&3(2x) - ???\end{align}$$

What do you do with the $12x$?
I know this seems extremely comical seeing as how easy it must be for you mathematical geniuses but... I'm still new to calculus and I really need help.

Answer 1

You have done it right for the first term $3x^2$ using the power rule and getting $3(2x)=6x$

For the second term, $-12x$, you have an easier way to get it .

Note that $ y=-12x$ is a the equation of a straight line and since derivatives are nothing but slopes we will get $y'= -12$ out of $y=-12x$

Of course you may use the power rule again considering that the power of $x$ in $-12x$ is $1$ and as the result we will have $ -12x^0 = -12$

Answer 2

HINT:

Given $k$ is a constant, derivative of $kx^n$ is given by,

$$\frac{d(kx^n)}{dx}=k(nx^{n-1})$$

Can you use the same rule to find derivative of $12x$?

Answer 3

You were solving it correctly. Recall the following property.

$$\frac{d}{dx} f(x)\pm g(x) = \frac{d}{dx} f(x)\pm\frac{d}{dx} g(x)$$

You can find the derivative of each individual term and add/subtract them to get the derivative of the given function.

$$\frac{d}{dx} 3x^2-12x = \frac{d}{dx} 3x^2-\frac{d}{dx} 12x$$

Now use the Power Rule for both individual parts.

$$\frac{d}{dx} ax^n = anx^{n-1} \implies \frac{d}{dx}3x^2 = 6x$$

$$\implies \frac{d}{dx} 12x = 12$$

Now, subtract.

$$\frac{dy}{dx} = 6x-12$$