3.5 Derivatives of trigonometric functions  (Page 2/3)

Using the product rule, we have

After simplifying, we obtain

By applying the quotient rule, we have

Simplifying, we obtain

To determine when the particle is at rest, set $s^{\prime }\left(t\right)=v\left(t\right)=0.$ Begin by finding $s^{\prime }\left(t\right).$ We obtain

so we must solve

The solutions to this equation are $t=\frac{\pi }{3}$ and $t=\frac{5\pi }{3}.$ Thus the particle is at rest at times $t=\frac{\pi }{3}$ and $t=\frac{5\pi }{3}.$

Start by expressing $\text{tan}x$ as the quotient of $\text{sin}x$ and $\text{cos}x:$

Now apply the quotient rule to obtain

Simplifying, we obtain

Recognizing that ${\text{cos}}^{2}x+{\text{sin}}^{2}x=1,$ by the Pythagorean theorem, we now have

Finally, use the identity $\text{sec}x=\frac{1}{\text{cos}x}$ to obtain

To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute

Thus the tangent line passes through the point $\left(\frac{\pi }{4},1\right).$ Next, find the slope by finding the derivative of $f\left(x\right)=\text{cot}x$ and evaluating it at $\frac{\pi }{4}\text{:}$

Using the point-slope equation of the line, we obtain

or equivalently,

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

In the first term, $\frac{d}{dx}\left(\text{csc}x\right)=\text{−}\text{csc}x\text{cot}x,$ and by applying the product rule to the second term we obtain

Therefore, we have