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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.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).
\[\mathop \limits_ \frac = \mathop \limits_ \frac = 6\mathop \limits_ \frac\] Note that we factored the 6 in the numerator out of the limit.
At this point, while it may not look like it, we can use the fact above to finish the limit.
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.
All derivatives of circular trigonometric functions can be found using those of sin(x) and cos(x).
The quotient rule is then implemented to differentiate the resulting expression.
See the Proof of Trig Limits section of the Extras chapter to see the proof of these two limits. Students often ask why we always use radians in a Calculus class. The proof of the formula involving sine above requires the angles to be in radians.
If the angles are in degrees the limit involving sine is not 1 and so the formulas we will derive below would also change.