Differencing x/4: What is the Rule?

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Im unsure on a very very basic differetiation i need for part of a question.

Quite simply - differentiate x/4. Thats it. Or x over any number - just never knew the rule. Is it simply 1? or 4? or 1/4?

I need to know for a question where iv to find the differential du for u = sin(x/y). Therefore i need to partially differentiate sin(x/y)
 
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Well, why not ponder the identity \frac{x}{4}=x*\frac{1}{4} for a while?
 
Since, I don't think this is a homework problem, I'll guide you and give you the answer.
There's a rule says that if k is a constant, then:
(k f(x))' = k f'(x). This can be proven by using the Product Rule.
Proof:
(k f(x))' = k' f(x) + k f'(x) = 0 f(x) + k f'(x) (the derivative of a constant is 0) = k f'(x) (Q.E.D)
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So applying that rule here, we have:
\left( \frac{x}{4} \right)' = \frac{1}{4} (x)' = \frac{1}{4}
You can also try the Product Rule or the Quotient Rule, both will work, but is a little bit longer.
Can you get this? :)
 
I'd already worked this out but thanks for that explanation - I am sure it'l help in the future. I am revising for exams - it was just a silly hurdle in a question - so no, not homework. Thanks again:smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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