Differentiate with respect to $x$ $$\arcsin \frac{2^{x+1}}{1+4^x}$$
I couldn't solve this problem.
Should I substitute anything or should I directly solve it?
Can you offer your assistance?
Thank you
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chndn
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1Did you use the Chain rule? – Amzoti Aug 25 '13 at 06:02
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nO OTHER easy method? – chndn Aug 25 '13 at 06:14
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Nothing fancy, in fact, it is a nice example of the chain rule. – Amzoti Aug 25 '13 at 06:18
2 Answers
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Chain rule is the way to go but you can make it easier this way: $$ \arcsin \Big(\dfrac{2x} {1+x^2}\Big) = 2\arctan x$$
So, this makes your expression $$\arcsin \frac{2^{x+1}}{1+4^x} = \arcsin \frac{2\cdot2^x}{1+(2^x)^2} = 2 \arctan (2^x) $$
Now use the chain rule.
Parth Thakkar
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2I can't tell that, come on! You gotta work it out yourself. If I tell that, what will you learn? – Parth Thakkar Aug 25 '13 at 06:41
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Imagine me throwing out the answer at you - you'd be at a loss buddy! One of my teachers always leads us to answer - rarely does he give away the answer - and that way of his has a taught me a lot. So, I won't (as far as possible) tell the answer straight away. – Parth Thakkar Aug 25 '13 at 07:10
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You need no fancy tricks: it’s just a rather tedious application of the chainrule. Since the outermost function is an arcsin, you first differentiate it as an arcsin:
$$\frac{d}{dx}\left(\arcsin\frac{2^{x+1}}{1+4^x}\right)=\frac1{\sqrt{1-\left(\frac{2^{x+1}}{1+4^x}\right)^2}}\cdot\frac{d}{dx}\left(\frac{2^{x+1}}{1+4^x}\right)\;.\tag{1}$$
The next step will be to differentiate the last factor in $(1)$ as a quotient. Eventually you’ll end up having to differentiate $2^{x+1}$ and $1+4^x$. You’ll also want to simplify that first factor on the righthand side of $(1)$.
Brian M. Scott
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@chndn: How far did you get? the next step should be to replace the derivative on the righthand side of $(1)$ with $$\frac{(1+4^x)\frac{d}{dx} \left(2^{x+1}\right) - 2^{x+1}\frac{d}{dx }\left(1+4^x\right)}{\left(1+4^x\right)^2};,$$ using the quotient rule. – Brian M. Scott Aug 25 '13 at 07:03