# Differentiating $e^{-2x}\sqrt{x^2 +1}$ and $\sin^{2}3x$

After scouting around I've had no luck in finding answers to my question which is relatively simple for alot of you.

Before jumping straight to the question I 'd like to clarify that I'm not only looking for an answer, I'm asking how you got that answer, if you don't mind ofcourse!

Differentiate with respect to x

$$e^{-2x}\sqrt{(x^{2} +1)}$$

If the first one is a bit too long to explain, feel free to answer and explain the following: $$\sin^{2}3x$$

• Have you studied product and chain rules for differentiation? – Nicky Hekster Sep 5 '14 at 7:14
• Yes I have studied those a while back and am somewhat familiar with them. Personal issue is that I've had to jump from Intermediate level to Advanced level in the subject, thus what I learned is basically 'half' of what there is to know – Juxhin Sep 5 '14 at 7:15
• OK, clear - because what you are facing here is a combination of those rules. Will give you hints. Ah - Did already wrote some down. – Nicky Hekster Sep 5 '14 at 7:17
• @NickyHekster Thanks alot nicky, appreciate it. I don't mind being thrown in the right direction and figuring out the rest for myself :-) – Juxhin Sep 5 '14 at 7:18

This should be your answers: $$-2e^{-2x}\sqrt{x^2+1} + e^{-2x}\frac{x}{\sqrt{x^2+1}}$$ $$6\cos(3x)\sin(3x),$$ which last expression can be written as $3\sin(6x)$.

• Thanks alot! Nicky do you mind if I ask how you've come to these conclusions or at least which formulas & rules you've used so that I can replicate these answers myself aswell? – Juxhin Sep 5 '14 at 7:24
• Use the formulas Did gave! For example put $u=e^{-2x}$ and $v=\sqrt{x^2+1}$. However, the u´s and v´s are not the same throughout. – Nicky Hekster Sep 5 '14 at 7:29
• Aha alright, starting to make sense now bit by bit, thanks alot! – Juxhin Sep 5 '14 at 7:29
• Before deciding my answers could I kindly ask what you meant exactly with the second part of your comment, "However, the u´s and v´s are not the same throughout" – Juxhin Sep 5 '14 at 7:37
• "I don't mind being thrown in the right direction and figuring out the rest for myself" Why make this impossible? – Did Sep 5 '14 at 7:51

Here are some useful rules. Consider some differentiable functions $u$ and $v$, then $$(uv)'=u'v+uv',\quad(\sqrt{u})'=\frac{u'}{2\sqrt{u}},\quad(\mathrm e^u)'=u'\mathrm e^u,\quad (u^2)'=2u'u.$$ These are special cases of the general fact that, if $F$ and $u$ are differentiable, then $$(F(u))'=u'\cdot F'(u).$$Can you solve the two examples in your question using these?

• Thanks! These rules you've posted certainly do help. I can't answer the last part of your post just yet but I will go and expand on what you've just told me. – Juxhin Sep 5 '14 at 7:23
• Thanks again for your help Did. I'm sorry I couldn't accept both answers as I wish I could've as the site doesn't allow so. – Juxhin Sep 5 '14 at 7:44
• Did you solve the examples yourself? Well, I guess you cannot anymore now that full formulas are on the page... so let me rephrase: did you manage to check the explicit solutions? – Did Sep 5 '14 at 7:50
• I still haven't manage to work it from A to Z but these posts did clear most of the fog in my head. What's left to do is some more research and try to grasp the concept behind the answers – Juxhin Sep 5 '14 at 7:54
• So, for all you know, the explicit formulas given to you for these two derivatives could be utter rubbish? (They are not.) The "concept behind the answers" is in my post, now what you need to do is to do the computations yourself (as they say, maths is not a spectator's sport...). – Did Sep 5 '14 at 8:05