Differentiating trigonometric functions questions with answers

Questions

1. Question 1sin, cos and tan of ax + b

   Given that $y = -\sin (2x - 1)$, where $x$ is in radians, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Let u be the angle, and differentiate it$u = 2x - 1$$\dfrac{du}{dx} = 2$
   2. Step 2: Differentiate −sin u with respect to u: which function does sin u become, and with which sign?$\dfrac{dy}{du} = -\cos u$
   3. Step 3: Chain rule: multiply, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = -\cos (2x - 1) \times 2$$= -2\cos (2x - 1)$
2. Question 2sin, cos and tan of ax + b

   Given that $y = 2\cos (5x + 3)$, where $x$ is in radians, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Let u be the angle, and differentiate it$u = 5x + 3$$\dfrac{du}{dx} = 5$
   2. Step 2: Differentiate 2cos u with respect to u: which function does cos u become, and with which sign?$\dfrac{dy}{du} = -2\sin u$
   3. Step 3: Chain rule: multiply, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = -2\sin (5x + 3) \times 5$$= -10\sin (5x + 3)$
3. Question 3Powers

   Given that $y = \sin^{4} x$, where $x$ is in radians, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Write y as a power of u, where u is the trigonometric function$u = \sin x$$y = u^{4}$
   2. Step 2: Differentiate each with respect to its own variable (chain rule for du/dx)$\dfrac{du}{dx} = \cos x$$\dfrac{dy}{du} = 4u^{3}$
   3. Step 3: Chain rule: multiply, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = 4\sin^{3} x \times \cos x$$= 4\sin^{3} x \cos x$
4. Question 4Powers

   Given that $y = \sin^{2} x$, where $x$ is in radians, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Write y as a power of u, where u is the trigonometric function$u = \sin x$$y = u^{2}$
   2. Step 2: Differentiate each with respect to its own variable (chain rule for du/dx)$\dfrac{du}{dx} = \cos x$$\dfrac{dy}{du} = 2u$
   3. Step 3: Chain rule: multiply, then write u in terms of x again$\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$$\dfrac{dy}{dx} = 2\sin x \times \cos x$$= 2\sin x \cos x$
5. Question 5Products

   Given that $y = x^{3}\tan 3x$, where $x$ is in radians, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Which rule do you need?Use the product rule.
   2. Step 2: Let u be the power of x and v the trigonometric function$u = x^{3}$$v = \tan 3x$
   3. Step 3: Differentiate each one (chain rule for v)$\dfrac{du}{dx} = 3x^{2}$$\dfrac{dv}{dx} = 3\sec^{2} 3x$
   4. Step 4: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = x^{3}\cdot 3\sec^{2} 3x + \tan 3x \cdot 3x^{2}$$= 3x^{3}\sec^{2} 3x + 3x^{2}\tan 3x$
6. Question 6Products

   Given that $y = x^{3}\cos x$, where $x$ is in radians, find $\dfrac{dy}{dx}$.

   Show the worked solution

   1. Step 1: Which rule do you need?Use the product rule.
   2. Step 2: Let u be the power of x and v the trigonometric function$u = x^{3}$$v = \cos x$
   3. Step 3: Differentiate each one (chain rule for v)$\dfrac{du}{dx} = 3x^{2}$$\dfrac{dv}{dx} = -\sin x$
   4. Step 4: u times dv/dx, plus v times du/dx$\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$$\dfrac{dy}{dx} = x^{3}(-\sin x) + \cos x \cdot 3x^{2}$$= -x^{3}\sin x + 3x^{2}\cos x$