Find the indicated derivative.h(t)=12t3+4t2t+3th(t)=\frac12t^3+\frac{4}{t^2}t+3\sqrt{t}h(t)=21t3+t24t+3th′(t)=h^{\prime}(t)=h′(t)=

3 2 t 2 − 8 t 3 + 3 2 t \frac32t^2-\frac{8}{t^3}+\frac{3}{2\sqrt{t}} 2 3 t 2 − t 3 8 + 2 t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 3

Find the indicated derivative. h ( t ) = 1 2 t 3 + 4 t 2 t + 3 t h(t)=\frac12t^3+\frac{4}{t^2}t+3\sqrt{t} h ( t ) = 2 1 t 3 + t 2 4 t + 3 t <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> h ′ ( t ) = h^{\prime}(t)= h ′ ( t ) =

In this problem, we're asked to find the derivative of the function h of t equals 1 half t cubed plus 4 over t squared plus 3 root t. So let's go ahead and get started in finding this derivative. Now looking at this function, I know that I wanna go ahead and rewrite a couple of these terms as power functions. So I have this one half t cubed. That is fine as is one half t cubed. It's really easy to see how we would use the power rule here. But for that second term, since I'm dealing with a fraction with my variable on the bottom, in order to make that variable on the top, I can just make that power negative. So the second term is 4 t to the power of negative 2, and that's gonna make it easier to apply the power rule. Now for my last term, I have 3 root t. Remember that the square root of anything is really just to the power of 1 half. So I can rewrite this as 3 times t to the power of 1 half, which just means I'm taking the square root there. Now it's much more easy to see for both of these terms that we can use the power rule. So let's go ahead and apply that power rule in finding the derivative. So let's start with that first term. I have 1 half t cubed. Now here, of course, wanna use the power rule, which means I'm gonna pull that 3 out to the front and decrease that exponent by 1. Now remember, because we already have this constant 1 half in front, I need to make sure that I'm accounting for that when I'm using the power rule. So when I pull that 3 out to the front, I wanna think of it as multiplying my original constant 1 half. So 3 times 1 half is going to give me 3 halves. So I have 3 halves times t to the power of 3 minus 1, which is just 2. Now moving on to my next term here, I have 4 t to the power of negative 2. Again, here, we wanna use the power rule. So I'm gonna pull that negative 2 out to the front and decrease that exponent by 1. Again, since I already have this 4 out front, I know that I'm taking that 4. I'm multiplying it by that negative 2 that I'm pulling out using the power rule. So that makes this a negative eight times t to the power of negative two minus one, which is negative 3. Now for my last term here, I have 3 times t to the power of 1 half. Again, using the power rule here, I'm gonna pull that one half to the front, decrease that power by 1, 3 times 1 half taking into account my original constant. And that one half I'm pulling out to the front gives me positive 3 halves times t to the power of 1 half minus 1, which is negative one half. Now this answer is already correct as is, but I'm gonna go ahead and rewrite this to more closely match how our original function was given to us. And this is typically what you're going to wanna do when presenting your final answers. Since our original function had 4 over t squared as opposed to 4 t to the power of negative 2, I'm going to go ahead and rewrite any of my negative powers back as a fraction. So let's go ahead and do that here. Now that first term is going to stay the same, three halves t squared. That's fine as is. And this is minus 8 over t cubed, taking away that negative, putting it back on the bottom of our fraction. Then for this next term, I have 3 over 2 and this is t to the power of negative one half. Now t to the power of 1 half we know is the square root of t. When it's a negative that just means it's on the bottom of the fraction. So I have t to the power of 1 half on the bottom which I can just rewrite as the square root of t on the bottom here. So this is my final answer in its final form. 3 has t squared minus 8 over t cubed plus 3 over 2 root t, and that's the derivative of my original function. Let me know if you have any questions, and I'll see you in the next one.

Find the indicated derivative. h ( t ) = 1 2 t 3 + 4 t 2 t + 3 t h(t)=\frac12t^3+\frac{4}{t^2}t+3\sqrt{t} h ( t ) = 2 1 t 3 + t 2 4 t + 3 t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> h ′ ( t ) = h^{\prime}(t)= h ′ ( t ) =

3 2 t 2 − 8 t 3 + 3 2 t \frac32t^2-\frac{8}{t^3}+\frac{3}{2\sqrt{t}} 2 3 t 2 − t 3 8 + 2 t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 3

t 2 − 2 t 3 + 1 t t^2-\frac{2}{t^3}+\frac{1}{\sqrt{t}} t 2 − t 3 2 + t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 1

3 2 t 2 − 8 t 3 + 3 2 t \frac32t^2-8t^3+\frac32\sqrt{t} 2 3 t 2 − 8 t 3 + 2 3 t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

3 2 t 3 − 8 t 2 + 3 2 t \frac32t^3-\frac{8}{t^2}+\frac32\sqrt{t} 2 3 t 3 − t 2 8 + 2 3 t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

3. Techniques of Differentiation

Basic Rules of Differentiation

Business Calculus

3. Techniques of Differentiation

Basic Rules of Differentiation

Struggling with Business Calculus?

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Multiple Choice

Find the indicated derivative.
$h(t)=\frac12t^3+\frac{4}{t^2}t+3\sqrt{t}$
$h^{\prime}(t)=$

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

$\frac32t^2-8t^3+\frac32\sqrt{t}$

$\frac32t^2-\frac{8}{t^3}+\frac{3}{2\sqrt{t}}$

$\frac32t^3-\frac{8}{t^2}+\frac32\sqrt{t}$

Verified step by step guidance

Step 1: Start by identifying the function h(t) = (1/2)t^3 + (4/t^2)t + 3√t. Rewrite the terms for clarity: h(t) = (1/2)t^3 + (4/t^2)t + 3t^(1/2).

Step 2: Differentiate each term of h(t) with respect to t using the power rule and the chain rule. For the first term, (1/2)t^3, apply the power rule: d/dt[(1/2)t^3] = (3/2)t^2.

Step 3: For the second term, (4/t^2)t, rewrite it as 4t^(-1) and differentiate: d/dt[4t^(-1)] = -4t^(-2).

Step 4: For the third term, 3t^(1/2), apply the power rule: d/dt[3t^(1/2)] = (3/2)t^(-1/2).

Step 5: Combine the derivatives of all terms to get the final derivative: h'(t) = (3/2)t^2 - 4t^(-2) + (3/2)t^(-1/2). Simplify further if needed.

3:59m

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