Find the derivative of the given function.f(t)=e3tt−2e−tf(t)=\frac{e^{3t}}{t-2e^{-t}}f(t)=t−2e−te3t

e3t(3t−8e−t−1)(t−2e−t)2\frac{$$e^{3t}(3t-8e$$^{-t}$$-1)}{(t-2e$$^{-t}$$)^2$$}(t−2e−t)2e3t(3t−8e−t−1)

Find the derivative of the given function. f ( t ) = e 3 t t − 2 e − t f(t)=\frac{e^{3t}}{t-2e^{-t}} f ( t ) = t − 2 e − t e 3 t

In this problem, we're asked to find the derivative of the function f of t is equal to e to the power of 3 t over t minus 2 times e to the power of negative t. Now because here we have a function dividing another function, we're going to have to use the quotient rule. If you have not yet learned the quotient rule, go ahead and circle back to this problem after you've gotten there. For the rest of you, let's get into this problem. We wanna find our derivative here, f prime of t. Now our memory tool for the quotient rule tells us low d high minus high d low over the square of what's below. So d just means we're taking the derivative. So our first term here on our numerator, low d high, so I'm taking t minus 2 times e to the negative t, so that's my low function, my function on the bottom, times the derivative of my high function. The derivative of e to the power of 3 t, this is going to be e to the power of 3 t multiplied using the chain rule here by the derivative of 3 t, which is just 3. Then minus so this is low d high minus high d low. So that's e to the power of 3 t times the derivative of this bottom function here. The derivative of t is just 1, so that's 1 minus 2 times e to the power of negative t. So if we go ahead and multiply here, this is 2 times e to the power of negative t. But because we need to apply the chain rule here, that means I need to multiply this by negative one. Now I can do this kind of all in one step so that I'm not gonna have to do a bunch of algebra in every single step. And I can go ahead and pull this negative one out front, multiplying by it there. That makes that negative a positive. So there's our numerator there. This is over the square of what's below. That's t minus 2e to the power of negative t squared. Make sure you don't forget to square that denominator there. Now we can do a little bit of simplification because there is quite a lot going on here. We can go ahead and pull e to the power of 3t out of the front or out of the top, I mean, because both of those terms have e to the power of 3t. Then inside of our parenthesis here, this is going to be 3 times t minus 2e to the negative t minus 1 +2e to the power of negative t. Now that denominator is staying the same for now. That is t minus 2e to the power of negative t squared. Now we want to simplify this as much as we can and I can see in the top here, if I go ahead and distribute this 3 and also distribute this negative, I will likely be able to combine some like terms. So we're going to focus in on just what's in these parentheses for now so we're not to rewrite everything over and over again. So if I distribute that 3, that is 3t minus 6e to the power of negative t. Then if I distribute this negative in my parentheses there, this is minus 1 minus 2e to the power of negative t. So I can combine these like terms. That's negative 6e to the power of negative t, negative two times e to the power of negative t. That gives me negative 8e to the power of negative t. And then I have a 3 t and I have a minus one. So this is 3t+3tminus1. So I can go ahead and rewrite this here. This is e to the power of 3t. And I'm going to keep 3t minus 1 in front. 3 t minus 1 minus 8 times e to the negative t. Then on the bottom there, keeping that denominator the same, that is t minus 2 e to the power of negative t squared. Now keep in mind that you may have simplified this slightly differently and you could still have the same exact answer and you would have the correct answer still. So be sure to double check and see if your answer, if it looks different now, if it actually does simplify to the same thing that I have here. So this is my final answer here, f prime of t, and my derivative is e to the power of 3 t times 3 t minus 1 minus 8 e to the power of negative t. All of this is over t minus 2 e to the power of negative t squared. Let me know if you have any questions here and let's keep going.

Find the derivative of the given function.f(t)=e3tt−2e−tf(t)=\frac{$$e^{3t}$$}{$$t-2e^{-t}$$}f(t)=t−2e−te3t

e3t(3t−8e−t−1)(t−2e−t)2\frac{$$e^{3t}(3t-8e$$^{-t}$$-1)}{(t-2e$$^{-t}$$)^2$$}(t−2e−t)2e3t(3t−8e−t−1)

3e3t1+2e−t\frac{$$3e^{3t}$$}{$$1+2e^{-t}$$}1+2e−t3e3t

(t−1)e3t(t−2e−t)2\frac{$$(t-1)e^{3t}$$}{$$(t-2e^{-t}$)^2$$}(t−2e−t)2(t−1)e3t

3t2e3t−1−8te2t−1−e3t(t−2e−t)2\frac{$$3t^2e$$^{3t-1}$$-8te^{2t-1}-e$$^{3t}$$}{(t-2e$$^{-t}$$)^2$$}(t−2e−t)23t2e3t−1−8te2t−1−e3t

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

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Multiple Choice

Find the derivative of the given function.
$f(t)=$\frac{e^{3t}$}{t-2e^{-t}}$

$$\frac{e^{3t}$(3t-8e^{-t}-1)}{(t-2e^{-t})^2}$

$$\frac{3e^{3t}$}{1+2e^{-t}}$

$$\frac{(t-1)e^{3t}$}{(t-2e^{-t})^2}$

$$\frac{3t^2e^{3t-1}$-8te^{2t-1}-e^{3t}}{(t-2e^{-t})^2}$

Verified step by step guidance

Identify the function for which you need to find the derivative: $ f(t) = \frac{e^{3t}}{t - 2e^{-t}} $. This is a quotient of two functions, so the quotient rule will be used.

Recall the quotient rule for derivatives: If $ f(t) = \frac{u(t)}{v(t)} $, then $ f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} $. Here, $ u(t) = e^{3t} $ and $ v(t) = t - 2e^{-t} $.

Find the derivative of the numerator $ u(t) = e^{3t} $. Using the chain rule, $ u'(t) = 3e^{3t} $.

Find the derivative of the denominator $ v(t) = t - 2e^{-t} $. The derivative is $ v'(t) = 1 + 2e^{-t} $ because the derivative of $ t $ is 1 and the derivative of $ -2e^{-t} $ is $ 2e^{-t} $.

Apply the quotient rule: $ f'(t) = \frac{3e^{3t}(t - 2e^{-t}) - e^{3t}(1 + 2e^{-t})}{(t - 2e^{-t})^2} $. Simplify the expression to get the final derivative.

6:32m

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Struggling with Calculus?

Find the derivative of the given function.f(t)=e3tt−2e−tf(t)=\(\frac{e^{3t}\)}{t-2e^{-t}}f(t)=t−2e−te3t

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