How do this differentiation question? (1 Viewer)

malcolm21 · Jun 13, 2015

Show that http://puu.sh/in9E1/e8d51dcb6b.png

Looks simple but im sure ive never learnt this

Silly Sausage · Jun 13, 2015

$2^x$ can be rewritten as $e^{xln2}$ .
To differentiate, let $u=xln2,\frac{du}{dx}=ln2$
Applying chain rule
$\frac{d}{dx}=2^x•ln2$

pomsky · Jun 13, 2015

It's a bit different to the usual because it's not possible to differentiate an exponential. Hence you will need to turn into a different form;
$2^x = e^{ln(2^x)}$ . This form works because e and ln are inverse functions.

When you have established this, you simply differentiate e as normal.
$\frac{d}{dx} e^{ln(2^x)} = \frac{d}{dx}e^{xln(2)}$ (using log laws)$$

$= ln(2)e^{xln(2)}$ (because ln(2) is simply a constant. Differentiating a constant (a) with respect to x; is equal to a (think $ \frac{d}{dx} 2x= 2)$ and when differentiating e: $ e^{ax} = ae^{ax}$ )

$= ln(2) \times 2^x$ (using $2^x = e^{ln(2^x)}$ )

And you get your answer

EDIT: Didn't see Silly Sausage's post- he explains it much more concisely than I do haha.

malcolm21 · Jun 13, 2015

^Thanks guys, I have another one though, the second part of this question: http://puu.sh/incIs/806cdb0ea0.png

rand_althor · Jun 13, 2015

malcolm21 said:
^Thanks guys, I have another one though, the second part of this question: http://puu.sh/incIs/806cdb0ea0.png

http://community.boredofstudies.org/13/mathematics-extension-1/337788/trig-question.html

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How do this differentiation question? (1 Viewer)

malcolm21

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Silly Sausage

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pomsky

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malcolm21

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rand_althor

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