Question

Differentiate following with respect to x.
a) e^2x

Answer 1

Hey there :) Remember your derivative for e^x?

Answer 2

Nope

Answer 3

\[\Large\rm y=e^x\]Take natural log of each side,\[\Large\rm \ln y=\ln e^x\]Apply log rule:\[\Large\rm \ln y=x \ln e\]\[\Large\rm \ln y=x\]Take derivative with respect to x,\[\Large\rm \frac{1}{y}y'=1\]Multiply by y,\[\Large\rm y'=y\]\[\Large\rm y'=e^x\]

Answer 4

Ok ok long story.. not so short...
derivative of e^x is e^x

Answer 5

Using this...

Answer 6

So we've found that taking the derivative of an exponential of base e, gives us the same thing back.
We need only apply the chain rule:\[\Large\rm y=e^{2x}\]\[\Large\rm y'=e^{2x}(2x)'\]

Answer 7

ok..

Answer 8

Chain rule tells us to multiply by the derivative of the inner function.
(Inner function was 2x in this case).

So what's our final answer?
Take the derivative of your 2x there.
(That's what the prime notation is telling us, that we still need to differentiate that part of it).

Answer 9

So, y=2x , dy/dx= 2

Answer 10

Good good good.\[\Large\rm y=e^{2x}\]\[\Large\rm y'=e^{2x}(2x)'\]\[\Large\rm y'=e^{2x}(2)\]It looks a little nicer if we bring the 2 to the front,\[\Large\rm y'=2e^{2x}\]

Answer 11

Boom, there we go.
Not too bad, right?

If you're still really confused on these, just practice. The chain rule is by far the most difficult of the derivative rules to master.

Answer 12

So, when v differentiate any exponent, we'll get back the same thing

Answer 13

Good question. The answer is no.

In general (We'll skip doing it the long way on this one) the derivative of an exponential is this:\[\Large\rm y=a^x\]\[\Large\rm y'=a^x(\ln a)\]

Answer 14

Notice that in our problem, yes we could have applied this rule,\[\Large\rm y=e^x\]\[\Large\rm y'=e^x(\ln e)\]But notice that log of e is simply 1, so we don't bother writing that step.

Answer 15

Okay

Answer 16

Thanks @zepdrix

Answer 17

np