Find the derivative of each function.f(t)=2t(t−3+t2/3)f(t)=2t(t^{-3}+t^{2/3})f(t)=2t(t−3+t2/3)

− 4 t − 3 + 10 3 t 2 / 3 -4t^{-3}+\frac{10}{3}t^{2/3} − 4 t − 3 + 3 10 t 2/3

Find the derivative of each function. f ( t ) = 2 t ( t − 3 + t 2 / 3 ) f(t)=2t(t^{-3}+t^{2/3}) f ( t ) = 2 t ( t − 3 + t 2/3 )

this problem, we're asked to find the derivative of the function f of t is equal to 2 t times t to the power of negative 3 plus t to the power of 2 thirds. So let's go ahead and jump right in here. We, of course, want to find our derivative f prime of t. And looking at my function here, I have 2 functions multiplying each other. I have 2 t multiplying the function t to the power of negative 3 plus t to the power of 2 thirds. So I'm going to use the product rule here. Now remember using the product rule, we know left d right plus a right d left is going to give us our derivative. So let's get started using that product rule. Now we're starting with that left hand function 2 t, leaving it as is, multiplying it by the derivative of our right function, that t to the power of negative 3 plus t to the power of 2 thirds. So taking the derivative of this function here using the power rule, I'm going to pull that negative 3 out to the front, decrease that power by 1 for that first term here, which is going to be negative 3 times t to the power of negative 3 minus 1, which is negative 4. Then I have this second term here again using the power rule, multiplying by that power 2 thirds, decreasing that power by 1. So this is going to give me 2 thirds times t to the power of 2 thirds minus 1, which is going to be a negative 1 third because, remember, I'm taking 2 thirds. I'm subtracting 1, which is the same thing as subtracting 3 over 3. So that's going to give me negative one over 3. So I found that first derivative here. Remember that we have a second term using our product rule. Now we have our right side function, leaving that as is. That's t to the power of negative 3 plus t to the power of 2 thirds. And then that's multiplying d left. That's the derivative of that left side function, which is a bit more simple here. It's 2 t. I know the derivative of 2 t is just 2. So from here, we can do a little bit of simplification to make this a little bit more simple. So I have this 2 t. I'm gonna go ahead and distribute this into this these parentheses, and I'm gonna do the same thing with my 2 over here. Now in doing that, I have 2 t times negative three t to the power of negative 4. That is going to give me a negative 6 t to the power of negative 3. And then I'm also multiplying here. This is going to give me 4 thirds t to the power of 2 thirds because this is t to the power of 1. This is t to the power of negative 1 third. Remember, when I multiply t to the power of something times t to the power of something else, I can just add those powers. Negative 1 third plus 1 is going to give me positive 2 thirds. So be really careful here when you're doing this algebra. Then for these next terms I'm just distributing that 2 in here. That's going to give me 2 t to the power of negative 3 plus 2 t to the power of 2 thirds. Now we can do some even more simplification here because we have some like terms. We have negative 6 t to the power of negative 3. We also have 2 t to the power of negative 3. So combining those two terms here, a negative 62, that's going to give me a negative 4 t to the power of a negative 3. Then I have another 2 terms that are alike. I have 4 thirds t to the power of 2 thirds, and I have 2 t to the power of 2 thirds. Now thinking of these in terms of fractions because I have this 4 thirds here, remember that 2 is the same thing as 6 thirds. So when I add these together, this is going to give me 10 over 3 t to the power of 2 thirds. And this is my final answer here. I can't combine anything else. Remember that you could always rewrite your negative exponents as putting that t to the power of 3 on the bottom. But since here we started out with a negative exponent in our problem statement, I'm going to leave that as is. So this is my final answer here. As you can see, sometimes the algebra is going to be the most challenging part of these derivative problems. So be sure to take your time and really pay close attention to that algebra. Let me know if you have any questions here, and I'll see you in the next one.

Find the derivative of each function. f ( t ) = 2 t ( t − 3 + t 2 / 3 ) f(t)=2t(t^{-3}+t^{2/3}) f ( t ) = 2 t ( t − 3 + t 2/3 )

− 4 t − 3 + 10 3 t 2 / 3 -4t^{-3}+\frac{10}{3}t^{2/3} − 4 t − 3 + 3 10 t 2/3

2 t − 3 + 2 t 2 / 3 2t^{-3}+2t^{2/3} 2 t − 3 + 2 t 2/3

− 3 t − 4 + 2 3 t − 1 / 3 -3t^{-4}+\frac23t^{-1/3} − 3 t − 4 + 3 2 t − 1/3

− 6 t − 3 + 4 3 t 2 / 3 -6t^{-3}+\frac43t^{2/3} − 6 t − 3 + 3 4 t 2/3

3. Techniques of Differentiation

Product and Quotient Rules

Business Calculus

3. Techniques of Differentiation

Product and Quotient Rules

Struggling with Business Calculus?

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Multiple Choice

Find the derivative of each function.
$f(t)=2t(t^{-3}+t^{2/3})$

$2t^{-3}+2t^{2/3}$

$-3t^{-4}+\frac23t^{-1/3}$

$-4t^{-3}+\frac{10}{3}t^{2/3}$

$-6t^{-3}+\frac43t^{2/3}$

Verified step by step guidance

Step 1: Rewrite the given function f(t) = 2t(t^{-3} + t^{2/3}) in a simplified form by distributing the 2t across the terms inside the parentheses. This gives f(t) = 2t * t^{-3} + 2t * t^{2/3}.

Step 2: Simplify each term by applying the laws of exponents. For the first term, 2t * t^{-3} becomes 2t^{1-3} = 2t^{-2}. For the second term, 2t * t^{2/3} becomes 2t^{1 + 2/3} = 2t^{5/3}. Thus, f(t) = 2t^{-2} + 2t^{5/3}.

Step 3: Differentiate each term of f(t) = 2t^{-2} + 2t^{5/3} using the power rule for derivatives, which states that d/dt[t^n] = n * t^{n-1}.

Step 4: For the first term, 2t^{-2}, the derivative is -2 * 2t^{-2-1} = -4t^{-3}. For the second term, 2t^{5/3}, the derivative is (5/3) * 2t^{(5/3)-1} = (10/3)t^{2/3}.

Step 5: Combine the results from Step 4 to write the derivative of f(t) as f'(t) = -4t^{-3} + (10/3)t^{2/3}.

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