2.3: Derivatives of Trigonometric Functions

Using the Product Rule, we have\[\begin{array}{rcl} f^{\prime}(x) & = & \dfrac{d}{dx}(5x^3) \cdot \sin x+\dfrac{d}{dx}(\sin x) \cdot 5x^3 \\[16pt] & = & 15x^2 \cdot \sin x+\cos x \cdot 5x^3. \\[16pt] \end{array}\nonumber\]After simplifying, we obtain\[f^{\prime}(x)=15x^2\sin x+5x^3\cos x.\nonumber\]

By applying the Quotient Rule, we have\[g^{\prime}(x)=\dfrac{(-\sin x)4x^2-8x(\cos x)}{(4x^2)^2}.\nonumber\]Simplifying, we obtain\[g^{\prime}(x)=\dfrac{-4x^2\sin x-8x\cos x}{16x^4}=\dfrac{-x\sin x-2\cos x}{4x^3}.\nonumber\]

To determine when the particle is at rest, set \(s^{\prime}(t)=v(t)=0\). Begin by finding \(s^{\prime}(t)\). We obtain\[s^{\prime}(t)=2 \cos t-1,\nonumber\]so we must solve\[2 \cos t-1=0\text{ for }0 \leq t \leq 2 \pi .\nonumber\]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 \(\tan x\) as the quotient of \(\sin x\) and \(\cos x\):\[f(x)=\tan x =\dfrac{\sin x}{\cos x}.\nonumber\]Now apply the Quotient Rule to obtain\[f^{\prime}(x)=\dfrac{\cos x\cos x-(-\sin x)\sin x}{(\cos x)^2}.\nonumber\]Simplifying, we obtain\[f^{\prime}(x)=\dfrac{\cos^2x+\sin^2 x}{\cos^2x}.\nonumber\]Recognizing that \(\cos^2x+\sin^2x=1\), by the Pythagorean Identity, we now have\[f^{\prime}(x)=\dfrac{1}{\cos^2x}.\nonumber\]Finally, use the identity \(\sec x=\frac{1}{\cos x}\) to obtain\[f^{\prime}(x)=\text{sec}^2 x.\nonumber\]

To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute\[f\left(\dfrac{\pi}{4}\right)=\cot\dfrac{\pi}{4}=1.\nonumber\]Thus the tangent line passes through the point \(\left(\frac{\pi}{4},1\right)\). Next, find the slope by finding the derivative of \(f(x)=\cot x\) and evaluating it at \(\frac{\pi}{4}\):\[f^{\prime}(x)=-\csc^2 x \quad \text{and} \quad f^{\prime}\left(\dfrac{\pi}{4}\right)=-\csc^2\left(\dfrac{\pi}{4}\right)=-2.\nonumber\]Using the point-slope equation of the line, we obtain\[y-1=-2\left(x-\dfrac{\pi}{4}\right)\nonumber\]or equivalently,\[y=-2x+1+\dfrac{\pi}{2}.\nonumber\]

To find this derivative, we must use both the Sum Rule and the Product Rule. Using the Sum Rule, we find\[f^{\prime}(x)=\dfrac{d}{dx}(\csc x)+\dfrac{d}{dx}(x\tan x ).\nonumber\]In the first term, \(\frac{d}{dx}(\csc x)=-\csc x\cot x \), and by applying the Product Rule to the second term we obtain\[\dfrac{d}{dx}(x\tan x )=(1)(\tan x )+(\sec^2 x)(x).\nonumber\]Therefore, we have\[f^{\prime}(x)=-\csc x\cot x +\tan x +x\sec^2 x.\nonumber\]

\[\begin{array}{rclcl} \dfrac{d}{dx} \left( \dfrac{\tan{(x)}}{1 + \cos{(x)}} \right) & = & \dfrac{\frac{d}{dx}\left( \tan{(x)} \right) \left[1 + \cos{(x)}\right] - \tan{(x)} \dfrac{d}{dx} \left( 1 + \cos{(x)} \right)}{\left[ 1 + \cos{(x)} \right]^2} & \quad & \left(\text{Quotient Rule}\right) \\[16pt] & = & \dfrac{\sec^2{(x)} \left[1 + \cos{(x)}\right] - \tan{(x)}\left[ -\sin{(x)} \right]}{\left[ 1 + \cos{(x)} \right]^2} & & \\[16pt] & = & \dfrac{\sec^2{(x)} \left[1 + \cos{(x)}\right] + \sin{(x)} \tan{(x)}}{\left[ 1 + \cos{(x)} \right]^2} & & \\[16pt] \end{array}\nonumber\]The tricky thing about trigonometric functions is the number of identities and the ultimate desire to simplify answers. It's often best to perform a few extra steps to see if any algebra or Trigonometry could simplify this result further.\[\begin{array}{rclcl} \dfrac{\sec^2{(x)} \left[1 + \cos{(x)}\right] + \sin{(x)} \tan{(x)}}{\left[ 1 + \cos{(x)} \right]^2} & = & \dfrac{\left[1 + \cos{(x)}\right] + \cos{(x)} \sin^2{(x)}}{\cos^2{(x)} \left[ 1 + \cos{(x)} \right]^2} & \quad & \left(\text{multiplying by }\dfrac{\cos^2{(x)}}{\cos^2{(x)}}\right) \\[16pt] & = & \dfrac{\left[1 + \cos{(x)}\right] + \cos{(x)} \left[1 - \cos^2{(x)}\right]}{\cos^2{(x)} \left[ 1 + \cos{(x)} \right]^2} & \quad & \left(\text{Pythagorean Identity}\right) \\[16pt] & = & \dfrac{\left[1 + \cos{(x)}\right] + \cos{(x)} \left[1 - \cos{(x)}\right]\left[1 + \cos{(x)}\right]}{\cos^2{(x)} \left[ 1 + \cos{(x)} \right]^2} & \quad & \left(\text{Difference of Squares}\right) \\[16pt] & = & \dfrac{\left[1 + \cos{(x)}\right]\left(1 + \cos{(x)} \left[1 - \cos{(x)}\right]\right)}{\cos^2{(x)} \left[ 1 + \cos{(x)} \right]^2} & \quad & \left(\text{factoring}\right) \\[16pt] & = & \dfrac{\cancel{\left[1 + \cos{(x)}\right]}\left(1 + \cos{(x)} \left[1 - \cos{(x)}\right]\right)}{\cos^2{(x)} \left[ 1 + \cos{(x)} \right]^{\cancelto{1}{2}}} & \quad & \left(\text{canceling}\right) \\[16pt] & = & \dfrac{1 + \cos{(x)} - \cos^2{(x)}}{\cos^2{(x)} \left[ 1 + \cos{(x)} \right]} & \quad & \left(\text{distributing}\right) \\[16pt] \end{array}\nonumber\]As you can see, our efforts to arrive at a more "simplified" form have not paid off. Any of these answers are equivalent, and none are really "simpler" than others; however, it is always advisable to perform a few extra steps like this to see if your results simplify (they often will).

Each step in the chain is straightforward:\[\begin{array}{rcl} y & = & \sin x \\[16pt] \dfrac{dy}{dx} & = & \cos x \\[16pt] \dfrac{d^2y}{dx^2} & = & -\sin x \\[16pt] \dfrac{d^3y}{dx^3} & = & -\cos x \\[16pt] \dfrac{d^4y}{dx^4} & = & \sin x \\[16pt] \end{array}\nonumber\]

Analysis

Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of \(\sin x\) equals \(\sin x\), so\[\dfrac{d^4}{dx^4}(\sin x)=\dfrac{d^8}{dx^8}(\sin x)=\dfrac{d^{12}}{dx^{12}}(\sin x)= \ldots =\dfrac{d^{4n}}{dx^{4n}}(\sin x)=\sin x\nonumber\]\[\dfrac{d^5}{dx^5}(\sin x)=\dfrac{d^9}{dx^9}(\sin x)=\dfrac{d^{13}}{dx^{13}}(\sin x)= \ldots =\dfrac{d^{4n+1}}{dx^{4n+1}}(\sin x)=\cos x.\nonumber\]

We can see right away that for the 74th derivative of \(\sin x\), \(74=4(18)+2\), so\[\dfrac{d^{74}}{dx^{74}}(\sin x)=\dfrac{d^{72+2}}{dx^{72+2}}(\sin x)=\dfrac{d^2}{dx^2}(\sin x)=-\sin x.\nonumber\]

First find \(v(t)=s^{\prime}(t)\).\[v(t)=s^{\prime}(t)=-\cos t .\nonumber\]Thus,\[v\left(\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}.\nonumber\]Next, find \(a(t)=v^{\prime}(t)\). Thus, \(a(t)=v^{\prime}(t)=\sin t\) and we have\[a\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}.\nonumber\]Since \(v\left(\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}<0\) and \(a\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}>0\), we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is traveling. Consequently, the particle is slowing down.