*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.

*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.*

\[\mathop \limits_ \frac = \frac\mathop \limits_ \frac = \frac\left( 1 \right) = \frac\] 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).

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.

Let θ be the angle at O made by the two radii OA and OB.

Since we are considering the limit as θ tends to zero, we may assume that θ is a very small positive number: The following derivatives are found by setting a variable y equal to the inverse trigonometric function that we wish to take the derivative of.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.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.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.Finding the derivatives of the inverse trigonometric functions involves using implicit differentiation and the derivatives of regular trigonometric functions.The diagram on the right shows a circle, centre O and radius r.\[\mathop \limits_ \frac = \mathop \limits_ \frac = 6\mathop \limits_ \frac\] 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.

## Comments Differentiation Of Trigonometric Functions Homework Answers

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