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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

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4 Answers 4

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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.

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  • $\begingroup$ 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 $\endgroup$ Commented Jun 27, 2014 at 14:40
  • $\begingroup$ @Andrew That root is just a constant, so you just have to apply the fact that $\dfrac{\mathrm d}{\mathrm dx}af(x)=a\dfrac{\mathrm d}{\mathrm dx}f(x)$. $\endgroup$ Commented Jun 27, 2014 at 14:43
  • $\begingroup$ @Andrew - Treat $\sqrt2$ the exact same way you just treated the $5$ in your example. It is also just a constant. $\endgroup$ Commented Jun 27, 2014 at 15:06
  • $\begingroup$ @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 $\endgroup$ Commented Jun 27, 2014 at 15:15
  • $\begingroup$ @Andrew - That's exactly right! $\endgroup$ Commented Jun 27, 2014 at 15:17
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Rewrite as $$ f(t)=\sqrt{2}t^{-7} $$ and use $$ \frac{d}{dt}t^\alpha = \alpha t^{\alpha-1}. $$

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

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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?

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  • $\begingroup$ I have edited now.. does this help you? $\endgroup$ Commented Jun 27, 2014 at 14:24
  • $\begingroup$ 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) $\endgroup$ Commented Jun 27, 2014 at 14:28
  • $\begingroup$ I have added one more step... can you complete that now? $\endgroup$ Commented Jun 27, 2014 at 14:35
  • $\begingroup$ is the answer sqrt(2)(-(7/t^8)), or sqrt(2)(-7t^-8)? $\endgroup$ Commented Jun 27, 2014 at 15:12
  • $\begingroup$ it is $-\sqrt{2}\cdot7^{-8}$ $\endgroup$ Commented Jun 27, 2014 at 16:43

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