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
$\endgroup$4 Answers
$\begingroup$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.
$\endgroup$ 5 $\begingroup$Rewrite as $$ f(t)=\sqrt{2}t^{-7} $$ and use $$ \frac{d}{dt}t^\alpha = \alpha t^{\alpha-1}. $$
$\endgroup$ $\begingroup$Hint: $$\rm\dfrac{d}{dx}ax^{b}=ab\,x^{\,b-1}.\tag{for all $\rm b\in\mathbb Z$}$$
$\endgroup$ $\begingroup$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?
$\endgroup$ 5
