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bettyboop8904
Can anyone explain how to differentiate exponential functions? I have the Calc 1 basic knowledge so I know the main differentiation rules (ie. 4x^2=8x) and I also know the addition, subtraction, multiplication, and division rules as well
You already have all the tools you need for differentiation. Do you have a specific example?
\[\large y=2^x\]Hmm so the variable is in the exponent position. That poses a problem. We can't apply the Power Rule that you may have learned at this point. We have to introduce logarithms in order to get the variable OUT of the exponent position.
I have e^5 and it says the answer is 0 but I don't understand why that is. and there are a couple others I don't get. Bc my book says that the exponential function is itself. Why wouldn't it be just f'(x)= e^5?
They're pulling a trick on you. See how e^5 contains no variable? It's just a fancy fancy looking CONSTANT, all dressed up in a suit.
lol @zepdrix has explained it better than i would
\[\large f(x)=e^5 \qquad \qquad \rightarrow \qquad \qquad f(x)=148.4\]See how the derivative is going to turn out? :) What happens when you take the derivative of a constant?
\[e^{2x}\div e^{4x ^{2}} + \cos(e ^{7x})\]
it approaches infinity?
that was me answering the post before i posted that big equation @zepdrix
i think i answered wrong lol
Nooo silly! :O That's one of those main derivative rules that you want to know at this point! :) A constant will give you 0 when you differentiate it. A derivative measures change. A constant is something that stays constant, doesn't change. So how much does `something that doesn't change` ... change? zerooooOooOO!
What do we need to do with the big messy problem? Take it's derivative?
That's right sorry, see the problem i have is answering questions that are put in to words haha but a lot of people struggle with that, exactly why i hate optimization problems = ( but yes derivative of a constant is 0. So why wouldn't it be 1 then? Isn't anything to the 0 power 1?
And differentiate it
taking the derivative and differentiating are the same thing right?
Yes, just fancy words ^^
ok that's what i thought lol
So I have the answer for the big equation from my neighbors notes but it doesn't make sense to me so that's why I want to ask about it = )
We can think about it like this, maybe this will help, maybe not :)\[\large f(x)=e^5\]See how there is no X value in the equation? Let's introduce an X.\[\large f(x)=(e^5)x^0\] \(x^0\) is just \(1\) right? So we're allowed to do that. We're just multiplying it by 1. Taking the derivative (applying the power rule to the x term) gives us,\[\large f'(x)=0(e^5)x^{0-1}\]We have a 0 coming down be multiplied by everything, that's going to turn the whole thing to 0. Maybe you're just getting confused by the e term. It contains no X's. We won't be trying to take it's derivative by applying the \(e^x\) rule.
Ok ok big question time c:
ok that makes sense = ) ty! and yes now to the big one lol
hold on lol cos is in the denominator with e to the 4x^2
\[\huge \frac{e^{2x}}{e^{4x^2}+\cos(e^{7x})}\]Ah ok so it is messy :) We'll need to apply the quotient rule.
\[\large \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\]
right that's what they did with it = ) that's where I got confused, not with the actual rule but u'
\[\huge \left(e^{2x}\right)'=e^{2x}(2x)'\]So we'll apply the e^x rule, but since we have more than just x in the exponent, we have to apply the chain rule. The prime on the brackets is to show that we need to take it's derivative still.
Remember how to do the chain rule? c:
\[\huge \left(e^{2x}\right)'=e^{2x}(2)\]Any confusion on that one? c:
oh wow i think i just had a mental click lmfao
lol XD those are always fun.
hold on let me look at the problem and answer one more time lol
omg I'm so dumb lol Idk if its bc how you explained it or if I was just looking at it blindly lol omg you were such a great help either way lol
Chain rule can be a little confusing at first. You're essentially making a 'copy' of the inner function and then taking its derivative, which can seem a little strange c:
but I do have another one lol it says \[\lim_{x \rightarrow \infty} (1.001)^{x}\] and it says the answer is infinity, why is that?
oh and then one more after that lol but i think it may be pretty easy haha
Just plug in a really big value for x and you'll see where it's heading! :) If x=99999\[\large 1.001^{99999}=2.55 \times 10^{43}\]
Since the base is LARGER than 1, and our exponent is positive, the value will explode.
What if you had,\[\large \lim_{x \rightarrow \infty} (.999)^x\]
so really big x values = really big y values
yah c: As x approaches infinity, y is approaching infinity. In that first case.
I used a calculator but the bigger x gets the closer y goes to 0 for the second part
In this second case, see how the BASE is less than 1? When we raise it to larger and larger powers it will get smaller and smaller and smaller. Imagine multiplying fractions. They get smaller! c: \[\large \dfrac{1}{2}\cdot \dfrac{1}{2} \quad = \quad \dfrac{1}{4}\]
so in that case it would go to -infinity
That's a good strategy with limits. If you're ever confused, just plug numbers in to see what happens.
Noooo :D An exponential can't go in the negative! :O
or wait it would go to 0 not -infinity right bc yeah what you just said haha
ok and so for the last one, well at least for now haha It says find the exponential function f(x)=Ca^{x} whose graph is given. It then has two problems... it shows you a graph from the exponent family and gives you two points on the graph, the first problem gives you the points: (1,6) and (3,24) I set the problem up as y=Ca^{x} and plugged 6 (the first y-value) in for y and then plugged 1 into the x. It says the answer is 3*2^{x}. So then I looked at the second point. I did the same thing as I did with the first pair but I couldn't see how you could figure it out without doing trial and error. The second problem gave the two points (2,2/9) and a y-intersect at 2. So I tried the same approach with the first pair and got 2*(1/3)^{x} as the answer but I'm not sure if that's correct
I feel the second answer I arrived at might be correct because a fraction with an exponent has a graph that is decreasing which the graph is decreasing in the picture of the second problem's graph @zepdrix
Ok simmer down :O Let's look at the first one a sec.
\[\large f(x)=Ca^x\]\[\large f(1)=6 \qquad \rightarrow \qquad 6=Ca^1\]\[f(3)=24 \qquad \rightarrow \qquad 24=Ca^{3}\]So this is where you got stuck on the first one?
We have to perform a little sneaky trick at this point! We'll DIVIDE our f values.\[\large \frac{f(3)}{f(1)} \qquad \rightarrow\qquad \frac{24}{6}=\frac{Ca^3}{Ca^1}\]We get a nice cancellation with the C's allowing us to solve for a.\[\large \frac{24}{6}=\frac{\cancel{C}a^3}{\cancel{C}a^1}\]
\[\large 4=a^2 \qquad \rightarrow \qquad a=2\]
ok that makes sense = ) is that a property or a rule?
It's this really fancy thing called "division" :3
or what i mean to say is how did you know to divide the points' equations hahahaha
Well here is something to think about. When you get to this point,\[\large 6=Ca^1\]\[\large 24=Ca^3\] You have a SYSTEM of equations. To be more specific, you have 2 equations and 2 unknowns. Since the number of unknowns does not exceed the number of equations given, we can find solutions for each unknown. Remember back to algebra, doing Substitution and Elimination? This is just another silly trick to add to your list. I'm not really sure what it's called though XD
We could have also done substition to get the same result though :)
like solve for C in one of the problems and then plug it into the other equation?
ok good i think i got it = ) thank you so much for all your help! = D