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I am an electronics student and today we were looking at the voltage output of a charged capacitor against time, which can be modelled in relation to R, C and e (I forget exactly how) but our teacher then differentiated it to find the gradient and it confused me because it was not how I thought differentiation worked wolfram alpha to the rescue?

I thought differentiation meant that ax^n went to axn^(n-1), but this clearly isn't what happened, so if someone could explain this it would be greatly appreciated!

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  • \$\begingroup\$ No, differentiation does not mean that ax^n went to axn. \$\endgroup\$ – immibis Oct 13 '16 at 23:23
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In this case 'e' is not a variable it is a constant: eulers number, ~2.718.... more commonly known as the base of natural log, ln().

The derivation rules for an exponential function such as a^x (where a is constant) state that the derivative for:
$$\frac{d}{dx}a^x = a^x \ln(a)$$

In your case, we have another constant to worry about, the 2 in front of the x. In which case, we copy down the constant to in front of the base (not bring it down as you were thinking) so:
$$\frac{d}{dx}a^{bx} = b a^{bx} \ln(a)$$
Where b is another constant.

So, there is one more point remaining, \$\ln(e) = 1\$. Relplacing \$a\$ with \$e\$ means that:
$$be^{bx}\ln(e) = b e^{bx}$$
by simply replacing b with 2, we get what wolfram alfa told you for the derivative of \$e^{2x}\$.

The main point: What you were thinking about differentiation is when the variable is the base of the exponent, not when the variable is in the exponent itself.

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  • \$\begingroup\$ @Dave Tweed, Thanks for that, I am fairly new to the forum, I had no Idea how to do the mathematics notation. \$\endgroup\$ – ambitiose_sed_ineptum Oct 14 '16 at 14:21

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