*This time we’re going to notice that it doesn’t really matter whether the sine is in the numerator or the denominator as long as the argument of the sine is the same as what’s in the numerator the limit is still one. \[\begin\mathop \limits_ \frac & = \left( \right)\left( \right)\ & = \left( \right)\left( \right)\ & = \left( 3 \right)\left( \right)\ & = \frac\end\] This limit almost looks the same as that in the fact in the sense that the argument of the sine is the same as what is in the denominator.*

*This time we’re going to notice that it doesn’t really matter whether the sine is in the numerator or the denominator as long as the argument of the sine is the same as what’s in the numerator the limit is still one. \[\begin\mathop \limits_ \frac & = \left( \right)\left( \right)\ & = \left( \right)\left( \right)\ & = \left( 3 \right)\left( \right)\ & = \frac\end\] This limit almost looks the same as that in the fact in the sense that the argument of the sine is the same as what is in the denominator.*

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.

Note that rules (3) to (6) can be proven using the quotient rule along with the given function expressed in terms of the sine and cosine functions, as illustrated in the following example.

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Before we start differentiating trig functions let’s work a quick set of limit problems that this fact now allows us to do. In fact, it’s only here to contrast with the next example so you can see the difference in how these work.

## Regent University Application Essay - Differentiation Of Trigonometric Functions Homework

In this case since there is only a 6 in the denominator we’ll just factor this out and then use the fact.

\[\mathop \limits_ \frac = \frac\mathop \limits_ \frac = \frac\left( 1 \right) = \frac\] Now, in this case we can’t factor the 6 out of the sine so we’re stuck with it there and we’ll need to figure out a way to deal with it.

To do this problem we need to notice that in the fact the argument of the sine is the same as the denominator ( both \(\theta \)’s).

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.

So, remember to always use radians in a Calculus class!

## Comments Differentiation Of Trigonometric Functions Homework

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