<p>f(x)=4∫2x sqrt(t^2-t)dt. Find f'(2).</p>
<p>How would you do this? I know the answer is 2sqrt(12) but I want to understand how to do it. Thanks in advance for any help.</p>
<p>Is that the integral from 4 to 2x?</p>
<p>Yes it is. Do you know how to do it?</p>
<p>at first i wasn’t sure how you could have x and y terms in an integral but i recognize this problem from the practice exam.</p>
<p>basically, plug in the 2x for every t, then multiply the whole radical by 2, which is the derivative of the upper bound 2x. so u get 2(sqrt((2x)^2-2x). then, u can now find f’(2). the concept behind this is FTC.</p>
<p>Fundemental Theorem of Calculus (I think?).</p>
<p>So to find f’(x), plug in 2x wherever there is a t, then multiply by the derivative of 2x (or whatver your upper bound is), which is just 2.</p>
<p>Then to find f’(2), just plug in 2 for x.</p>
<p>The derivative of a∫g(x) f(t)dt is f(g(x))<em>g’(x)
In your question: f(t)=sqrt(t^2-t), g(x)=2x, and g’(x)=2.
Your answer is then: f(g(x))g’(x) = sqrt((2x)^2-2x)</em>2 = 2<em>sqrt(4x^2-2x).
If you plug in 2 for x, you get 2</em>sqrt(4(2)^2-2<em>2) = 2</em>sqrt(12).</p>
<p>Alright. You need to know the second fundamental theorem of Calculus for this problem. </p>
<p>The bottom limit (4) does not come into play. When you want to find the derivative of an integral and the top limit has a variable in it, the derivative is the function itself with the top limit plugged in for the dependent variable (t in this case). </p>
<p>For example: the derivative of 0∫x f(t) is just f(x)</p>
<p>Now, the chain rule comes into play in your problem. You plug the 2x into your function to get:</p>
<p>sqrt(4x^2 - 2x), but you need to actually have 2sqrt(4x^2 - 2x) to reflect the chain rule (derivative of 2x is 2)</p>
<p>Now, just plug in 2: 2sqrt(16 -4) = 2sqrt(12)</p>
<p>Hope I helped. :)</p>
<p>EDIT: A bunch of people beat me while I was typing this up, but I think my explanation is the best. :p</p>
<p>Oh wow, I thought your b value was the 4 so I kept getting a negative answer, but yeah, it is really simple.</p>
<p>Thanks for the quick replies. I was stuck on this problem all day but you guys finally helped me figure it out.</p>