*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.*

At this point, while it may not look like it, we can use the fact above to finish the limit.Using the Pythagorean theorem and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.We're sorry, this computer has been flagged for suspicious activity.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.Using implicit differentiation and then solving for dy/dx, the derivative of the inverse function is found in terms of y.To convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting θ be y.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.\[\mathop \limits_ \frac = \frac\mathop \limits_ \frac = \frac\left( 1 \right) = \frac\] 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).

## Comments Differentiation Of Trigonometric Functions Homework Answers

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