Main content
Calculus 1
Course: Calculus 1 > Unit 2
Lesson 9: Derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x)- Derivatives of sin(x) and cos(x)
- Worked example: Derivatives of sin(x) and cos(x)
- Derivatives of sin(x) and cos(x)
- Proving the derivatives of sin(x) and cos(x)
- Derivative of 𝑒ˣ
- Derivative of ln(x)
- Derivatives of 𝑒ˣ and ln(x)
- Proof: The derivative of 𝑒ˣ is 𝑒ˣ
- Proof: the derivative of ln(x) is 1/x
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Derivative of 𝑒ˣ
The derivative of 𝑒ˣ is... well... 𝑒ˣ. This is a very special property lies at the heart of our work with exponential functions.
Want to join the conversation?
How would you use the "lim h-->0 f(x+h) - f(x)/h" strategy to solve for the derivative of e^x? I got 0/0, but it doesn't look like factoring or using conjugates work.(7 votes)
Factoring will work!
f(x)=e^x : this will be our original equation that we want to differentiate to achieve the general formula. As noted by this video, the general formula for this equation is the equation itself: e^x. Let's prove it using the general limit notation!
First, plug in (x) and (x+h) into the exponent.
f(x)= e^x
f(x+h)=e^(x+h)
Now that we have established the terms for the limit notation, we can continue:
lim h-->0 (e^(x+h) - e^x)/h
lim h-->0 (e^x times e^h - e^x)/h : bare with me, I know it's difficult to illustrate this but remind yourself that a^(b+c) is equivalent to a^b times a^c, as long as the base is the same, according the the exponential laws. Thus, e^(x+h) can be separated as e^x times e^h, since multiplying the two bases with the same exponent simply means to add the exponents together.
lim h-->0 [e^x(e^h - 1)]/h : factor out e^x, since this is the common factor. If you did not recognize this operation, you can set e^x = a
*** lim h-->0 (a times e^h - a)/h
*** lim h--->0 [a (e^h - 1)]/h : where e^x = a
lim h-->0 [e^x(e^h - 1)]/h : the coefficient, e^x does not depend on the denominator; it can be interpreted as (e^x)/1 times (e^h - 1)/h. Thus, move the e^x in front of the lim.
e^x times lim h-->0 (e^h - 1)/h : approximate the value of the limit by picking a number very close to zero to input into 'h' value. Let's choose .0001 (you can choose .01, .001, .0000000000000000000001...but you will find that as h gets closer to zero, the value of the limit gets more accurate).
e^x times lim h-->0 (e^0.0001 - 1)/0.0001 : the value of the limit is 1
e^x times 1
f'(x)= e^ x : this proves that the derivative (general slope formula) of f(x)= e^x is e^x, which is the function itself. In other words, for every point on the graph of f(x)=e^x, the slope of the tangent is equal to the y-value of tangent point. So if y= 2, slope will be 2. if y= 2.12345, slope will be 2.12345(29 votes)
Wouldn't e^x fall under the power rule? So why does it behave differently?(4 votes)
The power rule concerns functions of the form x^n, where the base is the variable and the exponent is a constant.
In the case of e^x, the base is constant and the exponent is variable, so it isn't the same thing.(28 votes)
But shouldn't be the derivative of e^x = xe^(x-1) using Power Rule?(1 vote)
No the power rule only works for a constant in the exponent, when the exponent is the input to the function or some function of the input to the function, you cannot use the power rule. Hope this helps!(18 votes)
How did he got the slope so fast??(2 votes)
He's just eyeballing it; it looks pretty close to the values he says.(5 votes)
- Does that mean, at y = infinity, the graph will be a vertical line and thus, the slope of e^x at y = infinity will be undefined?(4 votes)
yuh, i guess if ur talking about y=infinity that's the case. The slope for that equation will approach infinity as x->oo. same thing happens with equations like x^2(1 vote)
- At2:03, Sal says "the derivative with respect to X". Can we have a derivative in respect to another variable, say, t (time) for a function of time (f(t)) like speed in a race. Would the derivative be expressed as d/dt f(t)? Is this possible?(3 votes)
- You are exactly right. You can take the derivative with respect to any variable and it will look just like that.
Also, like you said, the variable should match the function. If you wanted to take d/dx f(t), it would be 0, because f(t) does not depend on x or change as x changes.(1 vote)
- Is this function, e^(x) really differentiable?if yes then how. My math teacher says it isn't actually differentiable.(0 votes)
Yes, the function e^x is differentiable, and its derivative is e^x.
Have a blessed, wonderful day!(6 votes)
What would the derivative of e^(-5x) be then? Would it be the same thing?(1 vote)- You apply the chain rule, first saying e^(-5x) = e^u*, where u = -5x and du/dx = -5. Differentiating, you get du/dx e^u, which equals -5e^(-5x).(4 votes)
- Is this true as well for f(x)= 2^x? I dont know how to find the derivative of this. My professor wants us to use the definition of a derivative to solve.(2 votes)
- No. d/dx(2^x) is not equal to 2^x. There are other videos that go into more detail about differentiation of exponential functions.(1 vote)
- How would you differentiate e^-x?(1 vote)
It uses the chain rule, which is done in further lessons.(1 vote)
2:03Video transcript
- [Instructor] What we
have right over here is the graph of Y is equal to E to the X and what we're going to know
by the end of this video is one of the most
fascinating ideas in calculus and once again it reinforces the idea that E is really this
somewhat magical number. So we're gonna do a little
bit of an exploration. Let's just pick some points on this curve of Y is equal to E to the X and think about what the
slope of the tangent line is or what the derivative looks like and so let's say when Y is equal to one or when E to the X is equal to one, this is the case when X is equal to zero. Well, the slope of the tangent line looks like it is one,
which is curious because that's exactly the value of
the function at that point. What about when E to the X is
equal to two right over here? Well here, let me do it in another color, the slope of the tangent
line sure looks pretty close, sure looks pretty close to two. What about when E to
the X is equal to 1/2? So that's happening right about here. Well, it sure looks like the
slope of the tangent line is about 1/2. We could try what happens when E to the X is equal to five? Well, the slope of the tangent line here sure does look pretty close, sure does look pretty close to five and so just eyeballing it, is it the case that the
slope of the tangent line of E to the X is the same
thing, is E to the X? And I will tell you and
this is an amazing thing that that is indeed true, that if I have some function, F of X, that is equal to E to the X and if I were to take
the derivative of this, this is going to be equal
to E to the X as well or another way of saying it, the derivative with
respect to X of E to the X is equal to E to the X and that is an amazing thing. In previous lessons or courses, you've learned about ways to define E and this could be a new one. E is the number that where
if you take that number to the power of X, if
you define a function or expression as E to the X, it's that number where if you
take the derivative of that it's still going to be E to the X. And what you're looking here, this curve, it's a curve where the
value that's Y value at any point is the same as
the slope of the tangent line. If that doesn't strike you as mysterious and magical and amazing just yet, it will. Maybe tonight you'll wake up
in the middle of the night and you'll realize just what's going on. Now, some of you might be
saying okay, this is cool, you're telling me this, but
how do I know it's true? In another video, we will do the proof.