# How to find the derivative of a fraction?

So I have this fancy problem I've been working on for two days:

I need to find two things:

1) $f'(t)$
2) $f'(2)$

I have tried plugging it into the definition of a derivative, but do not know how to solve due to its complexity.

Here is the equation I am presented:

If $f(t) = \sqrt{2}/t^7$ find $f'(t)$, than find $f'(2)$.

How do I convert this problem into a more readable format? (no fractions or division), otherwise, how do I complete it with the fractions?

Thanks

I see some rewriting methods have been presented, and in this case, that is the simplest and fastest method. But it can also be solved as a fraction using the quotient rule, so for reference, here is a valid method for solving it as a fraction.

Let $f(x) = \frac{\sqrt 2}{t^7}$

Let the numerator and denominator be separate functions, so that $$g(x) = \sqrt2$$ $$h(x) = t^7$$

So $$f(t) = \frac{g(t)}{h(t)}$$

The quotient rules states that $$f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{h^2(t)}$$

Using $$g'(t) = \frac{d}{dt}\sqrt2 = 0$$ $$h'(t) = \frac{d}{dt}t^7 = 7t^6$$

we get, by plugging this into the quotient rule: $$f'(t) = \frac{0\cdot t^7 - \sqrt2\cdot7t^6}{t^{14}}$$

Simplifying this gives us $$\underline{\underline{f'(t) = -\frac{7\sqrt2}{t^8}}}$$

This is also the same as the result you should get by rewriting $$f(t) = \frac{\sqrt2}{t^7} = \sqrt2 \cdot t^{-7}$$ and using the power rule.

• I understand how to use the power rule. I just don't understand how it applies when there is a root in front. If that makes sense. So I know that 5x^3 = 10x^2 etc. But I don't understand how to approach sqrt(2) * t ^ -7 Jun 27, 2014 at 14:40
• @Andrew That root is just a constant, so you just have to apply the fact that $\dfrac{\mathrm d}{\mathrm dx}af(x)=a\dfrac{\mathrm d}{\mathrm dx}f(x)$. Jun 27, 2014 at 14:43
• @Andrew - Treat $\sqrt2$ the exact same way you just treated the $5$ in your example. It is also just a constant.
– Alec
Jun 27, 2014 at 15:06
• @Aleksander - So would the result than be -7(sqrt(2))t^-8? Or am I still missing a step? I think I'm getting it Jun 27, 2014 at 15:15
• @Andrew - That's exactly right!
– Alec
Jun 27, 2014 at 15:17

Rewrite as $$f(t)=\sqrt{2}t^{-7}$$ and use $$\frac{d}{dt}t^\alpha = \alpha t^{\alpha-1}.$$

Hint: $$\rm\dfrac{d}{dx}ax^{b}=ab\,x^{\,b-1}.\tag{for all \rm b\in\mathbb Z}$$

Hint : For any $a\in \mathbb{R}$ $$\frac{d}{dx}ax^n=anx^{n-1}$$

With this we have :

$$f(t) = \sqrt{2}t^{-7}\Rightarrow f'(t)=\sqrt{2}(-7t^{-7-1})$$

Can you Complete this?

• it is $-\sqrt{2}\cdot7^{-8}$