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# Finding derivative with fundamental theorem of calculus: x is on both bounds

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)

## Video transcript

so let's see if we can take the derivative of this expression right over here if we can find capital F prime of X and once again it looks like you might be able to use the fundamental theorem of calculus the big giveaway is that you're taking the derivative of a definite integral that gives you a function of X but here I have X on both the upper and the lower boundary and the fundamental theorem of calculus at least for them what we've seen it it's when we have X is only on the upper boundary then of course it's an x squared and but we've seen examples of that already when we use the chain rule to do it but how can we break this up and put this in a form that's a little bit closer to what we're familiar with when we apply the fundamental theorem of calculus and to realize that we really just have to attempt to graph what this is representing so let's say that this is our lower case f of X or I should say f of T so let's get called this lower case F of T and let's graph it over the interval between X and x squared so let's say this is my Y axis this is my t axis and let's say that this right over here is y is equal to f of T I'm drawing it generally I don't know what this exactly looks like and we're going to talk about the interval between x and x squared so if we're going to talk about the interval between X which is right over here so lower bound so x and x the square it's the lower bound at least for this definite integral we don't know for sure it depends on what x you choose on which one is actually smaller but let's just say that right for the sake of visualizing we'll draw X right over here and we will draw will draw x squared right over here x squared right over here so this whole expression this entire definite integral is essentially asking for is essentially representing the this entire area the entire area under under the curve but what we could do is introduce a constant that's someplace in between X and x squared let's say that constant is C and break this area into two different areas with C as the divisor so that same exact whole area we can now write it as two separate integrals so one integral that represents this area right over here and then another integral that this area right over there and we just say C is some constant between X and x squared well how can we denote this area in purple well that's going to be so this thing is going to be equal to the sum of these two areas the purple area is going we can show is the definite integral from X to C of our function of T cosine T over T DT and then to that we're going to add the green area and then we'll get the original area so the green area for the green area our lower bound of integration is now our constant C and our upper bound of integration is x squared it's going to be of cosine T D over T cosine T over T DT and this is a form where if we if we know how to apply the chain rule we can apply the fundamental theorem of calculus and this is almost in a form we're used to seeing it where the X is the upper bound now what we already know what happens we can swap these two bounds but it will just be the negative of that integral so this is going to be equal to let me rewrite it the negative the negative of the definite integral from C to X of cosine T over T DT and then we have plus the definite integral that goes from C to x squared of cosine T cosine T over T DT so all we've done is we've rewritten this thing in a way that it's we're used to applying the fundamental theorem of calculus so if we want to find F prime F prime of X well applying the derivative operator over here we're going to have a negative up front it's going to be equal to negative cosine X over X cosine X over X once again just the fundamental theorem of calculus and then plus we're first going to take the derivative of this thing with respect to x squared and that's going to give you cosine of x squared over x squared wherever you saw t you replace it with an x squared and then you're going to multiply that times the derivative of x squared with respect to X so that's just going to be the derivative of x squared with respect to X is just 2x and we're done we just need to simplify this thing so all of this is going to be equal to negative cosine x over X plus well this is going to cancel out with just one of those plus two cosine x squared over X and I guess we could simplify it even more as being equal to and we can swap these everything over x2 cosine of x squared minus cosine of X and we are done