Is this very simple derivative correct?

Question

I apologize for such a very easy question, but I'm really rusty. Is this very simple derivative correct?

$$ \frac{{\rm d} }{{\rm d}x} 2^{(x+1)^2} = 2^{(x+1)^2}\ln(2)$$

Or would it be like this:

$$ \frac{{\rm d} }{{\rm d}x} 2^{(x+1)^2} = 2(x+1) \cdot 2^{(x+1)^2} \cdot \ln(2)$$

I am using this nice pdf to remember the rules, but I am not sure about this case. Thank you!

@vadim123 thank you! I was doubting about it. It has been a long time without touching derivatives. — iadvd, Oct 04 '16 at 00:14
$2^{(x+1)^2}=e^{ln2\cdot(x+1)^2}$; then apply chain rule: $(e^{f(x)})'=e^{f(x)}f'(x)$ — Alex, Oct 04 '16 at 00:15
@Alex Thank you! if you have time and you could add that explanation in an answer I would gladly accept it. — iadvd, Oct 04 '16 at 00:16
@iadvd In the future, if you just want to verify a derivative or integral or other calculation, you can use WolframAlpha. — , Oct 04 '16 at 00:18

Alex · Accepted Answer · 2016-10-04T01:04:01.423

2

In calculus, I teach my students this substitution.

$2^{(x+1)^2}=e^{\ln 2\cdot(x+1)^2}$

Then remembering derivatives of things with $x$ in the power are quite simple: just apply chain rule, which states

$(e^{f(x)})'=e^{f(x)}f'(x)$

In this case:

($e^{\ln 2\cdot(x+1)^2})'=\ln 2\cdot 2(x+1)\cdot e^{\ln 2\cdot(x+1)^2}=\ln 2\cdot 2(x+1)\cdot 2^{(x+1)^2}$

The last step is just back-substituting the original function back into the result.

Note that this same rule also works when differentiating functions with $x$ both in the bottom and the power:

$(x^x)' = (e^{x\cdot \ln x})' = (1\cdot \ln x + x\cdot 1/x)e^{x\cdot \ln x} = (\ln x+1)x^x$

edited Oct 04 '16 at 01:04

answered Oct 04 '16 at 00:22

Alex

2,449

thank you very much for taking the time to answer! The reason for this question was asking this this new one: http://math.stackexchange.com/questions/1952774/it-is-possible-to-apply-a-derivative-to-both-sides-of-a-given-equation-and-maint – iadvd Oct 04 '16 at 00:36
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In my opinion, this is the absolutely best way to teach people how to do these types of derivatives.. Remembering the formulas for $log_a(x)$ and $a^x$ without $a=e$ is missing a good opportunity to show why $ln(x)$ rules and other tricks are useful. – mike van der naald Oct 04 '16 at 00:56

Claude Leibovici · Answer 2 · 2016-10-04T03:36:14.930

Another trick which is useful when you face products, quotients, powers, ... : it is logarithmic differentiation.

Let us consider your case $$y=2^{(x+1)^2}\implies \log(y)=(x+1)^2\,\log(2)$$ Differentiate both sides $$\frac{y'}y=2(x+1)\log(2)\implies y'=2(x+1)\log(2)\,2^{(x+1)^2}$$

Let do the same for $$y=f(x)^{g(x)}\implies\log(y)=g(x)\log(f(x))$$Differentiate both sides $$\frac{y'}y=\frac{g(x) f'(x)}{f(x)}+g'(x)\log (f(x)) $$ $$y'=f(x)^{g(x)-1} \big(g(x) f'(x)+f(x) \log (f(x)) g'(x)\big)$$

Is this very simple derivative correct?

2 Answers2