Differentiation Of Trigonometric Functions Homework

Differentiation Of Trigonometric Functions Homework-17
We’ll start this process off by taking a look at the derivatives of the six trig functions. The remaining four are left to you and will follow similar proofs for the two given here.Before we actually get into the derivatives of the trig functions we need to give a couple of limits that will show up in the derivation of two of the derivatives.The formulas below would pick up an extra constant that would just get in the way of our work and so we use radians to avoid that.

We’ll start this process off by taking a look at the derivatives of the six trig functions. The remaining four are left to you and will follow similar proofs for the two given here.Before we actually get into the derivatives of the trig functions we need to give a couple of limits that will show up in the derivation of two of the derivatives.The formulas below would pick up an extra constant that would just get in the way of our work and so we use radians to avoid that.

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All we need to do is multiply the numerator and denominator of the fraction in the denominator by 7 to get things set up to use the fact. \[\begin\mathop \limits_ \frac & = \frac\\ & = \frac\\ & = \frac\\ & = \frac\end\] This limit looks nothing like the limit in the fact, however it can be thought of as a combination of the previous two parts by doing a little rewriting.

First, we’ll split the fraction up as follows, \[\mathop \limits_ \frac = \mathop \limits_ \frac\frac\] Now, the fact wants a \(t\) in the denominator of the first and in the numerator of the second.

Therefore, after doing the change of variable the limit becomes, \[\mathop \limits_ \frac = \mathop \limits_ \frac = 1\] The previous parts of this example all used the sine portion of the fact.

However, we could just have easily used the cosine portion so here is a quick example using the cosine portion to illustrate this.

So we need to get both of the argument of the sine and the denominator to be the same.

We can do this by multiplying the numerator and the denominator by 6 as follows.To see that we can use the fact on this limit let’s do a change of variables.A change of variables is really just a renaming of portions of the problem to make something look more like something we know how to deal with.Note that we didn’t really need to do a change of variables here.All we really need to notice is that the argument of the sine is the same as the denominator and then we can use the fact.This is easy enough to do if we multiply the whole thing by \(\) (which is just one after all and so won’t change the problem) and then do a little rearranging as follows, \[\begin\mathop \limits_ \frac & = \mathop \limits_ \frac\frac\frac\ & = \mathop \limits_ \frac\frac\ & = \left( \right)\left( \right)\end\] At this point we can see that this really is two limits that we’ve seen before.Here is the work for each of these and notice on the second limit that we’re going to work it a little differently than we did in the previous part.If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.With this section we’re going to start looking at the derivatives of functions other than polynomials or roots of polynomials.This is not the problem it appears to be once we notice that, \[\frac = \frac\] and then all we need to do is recall a nice property of limits that allows us to do , \[\begin\mathop \limits_ \frac & = \mathop \limits_ \frac\ & = \frac\ & = \frac\end\] With a little rewriting we can see that we do in fact end up needing to do a limit like the one we did in the previous part.So, let’s do the limit here and this time we won’t bother with a change of variable to help us out.

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