# Differentiating $7x$

I have the question "Differentiate with respect to $x, 7x$."

I use the rule $nx^{n-1}$ with n being the power.

So using this rule I got since the power is $1$,

$1-1 = 0$ and so $7x^0$ should be $1$ because anything to the power $0$ is $1$.

But the solutions say that the answer should be $7$ why is this ?

• Because exponentiation has precedence over mutiplication. – user228113 Dec 4 '16 at 18:01
• It's $7(x^0)$, not $(7x)^0$. – tilper Dec 4 '16 at 18:03
• Ah I see so it's 7 x 1 = 7. Thanks (: – Dan Dec 4 '16 at 18:06
• Alternatively, when $f(x)=7x$ we have the difference quotient $$\frac{f(x+h)-f(x)}{h}=\frac{7(x+h)-7x}{h}=\frac{7h}{h}=7.$$ Since this doesn't change as $h\to 0$, the derivative of $f(x)$ is $f'(x)=7$. – Semiclassical Dec 4 '16 at 18:10

To apply the power rule here, you must recognize that

$$\frac d{dx}7x=7\left(\frac d{dx}x\right)$$

And that

$$\frac d{dx}x=1\times x^0=1$$

So that our original derivative is

$$\frac d{dx}7x=7\times1=7$$

You're using the power rule incorrectly. If you were to use the power rule for $f(x) = 7x$, then you'd get $1*7*x^{1-1} = 1 * 7 * 1 = 7$; however, a more simple rule is the constant multiple rule which states $\forall n \not\in X$, $\frac{d}{dx}(nx) = n$. Where $X$ is the domain of the function $f : X \to Y$

$$\frac{d[af(x)]}{dx}=a\times \frac{d[f(x)]}{dx}$$

Here $$a=7$$ and $$f(x)=x^1$$

$$\frac{d[7\times x^1]}{dx}=7\times \frac{d[x^1]}{dx}$$

$$=7\times 1(x)^{1-1}$$

$$\frac{d[7\times x^1]}{dx}=7\times 1(x)^0$$

$$\frac{d[7\times x^1]}{dx}=7$$