Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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 – Andrew Jun 27 '14 at 14:40
@Andrew That root is just a constant, so you just have to apply the fact that $\dfrac{\mathrm d}{\mathrm dx}af(x)=a\dfrac{\mathrm d}{\mathrm dx}f(x)$. – Hakim Jun 27 '14 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 '14 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 – Andrew Jun 27 '14 at 15:15
@Andrew - That's exactly right! – Alec Jun 27 '14 at 15:17

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

share|cite|improve this answer

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

share|cite|improve this answer

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?

share|cite|improve this answer
I have edited now.. does this help you? – Praphulla Koushik Jun 27 '14 at 14:24
Yes, I am just not sure of the operations after the exponent is placed infront of the sqrt(2). Do I multiply the 2 by -7, or 2^(1/2) by -7. Because 2^(1/2) == sqrt(2) – Andrew Jun 27 '14 at 14:28
I have added one more step... can you complete that now? – Praphulla Koushik Jun 27 '14 at 14:35
is the answer sqrt(2)(-(7/t^8)), or sqrt(2)(-7t^-8)? – Andrew Jun 27 '14 at 15:12
it is $-\sqrt{2}\cdot7^{-8}$ – Praphulla Koushik Jun 27 '14 at 16:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.