I need help with the following derivatives

y= (e^1/x) / x^2

f(t) = t ln(5t)

thanks for your help

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- Apr 13th 2010, 06:05 PMtim_manniredifferentiation
I need help with the following derivatives

y= (e^1/x) / x^2

f(t) = t ln(5t)

thanks for your help - Apr 13th 2010, 06:17 PMdrumist
Could you show us what exactly is giving you trouble? What do you get when you attempt them?

- Apr 13th 2010, 06:18 PMharish21
- Apr 14th 2010, 02:45 AMtim_mannire
Using the rules you suggested i got

dy/dx = ((x^2)(e^1/x) - (e^1/x)(2x)) / x^4

f ' (t) = (1/5t) (t)

are they correct? - Apr 14th 2010, 03:50 AMe^(i*pi)
For dy/dx it appears you've not differentiated $\displaystyle e^{1/x}$ correctly

It's easier if you sub $\displaystyle u = \frac{1}{x}$ and then use the chain rule.

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For the second one remember the chain rule

$\displaystyle \frac{d}{dx} \ln [f(x)] = \frac{f'(x)}{f(x)}$ - Apr 14th 2010, 05:25 AMtim_mannire
thanks for your help. I have tried what you suggested but I just don't understand (Headbang). A complete working of the question using the rules would help me to understand.

thank you - Apr 14th 2010, 05:38 AMe^(i*pi)
$\displaystyle y = \frac{e^{1/x}}{x^2}$

For this we need to use the quotient rule which is: $\displaystyle y' = \frac{vu' - uv'}{v^2}$

In your case $\displaystyle u = e^{1/x}$ and $\displaystyle v = x^2$

I will differentiate the numerator as that's what seems to be the issue then you can try using my value of $\displaystyle u'$ in the quotient rule. FYI my final answer is $\displaystyle \frac{dy}{dx} = \frac{e^{1/x}(2x-1)}{x^4} = -\frac{e^{1/x}(1-2x)}{x^4}$

To find u' (du/dx) we need to work out $\displaystyle \frac{d}{dx} e^{1/x}$.

For simplicity and clarity's sake let $\displaystyle t = \frac{1}{x}$. From the chain rule we know that $\displaystyle \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx}$

Due to this substitution we can rewrite u as $\displaystyle u = e^t$.

Differentiating this function is easy: $\displaystyle \frac{du}{dt} = e^t$

Similarly 1/x is also easy to differentiate: $\displaystyle \frac{dt}{dx} = -\frac{1}{x^2}$

From the chain rule above: $\displaystyle u' = \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx} = e^t \cdot \frac{1}{x^2}$

Since $\displaystyle t = \frac{1}{x}$ we get $\displaystyle u'$ in terms of x:

Therefore $\displaystyle u' = -\frac{1}{x^2}\,e^{1/x}$ - Apr 14th 2010, 08:43 AMdrumist