Math 10350 - Calculus Derivatives (Product, Quotient, Higher, Trig Functions)

Product Rule & Quotient Rule

Product Rule

  • Definition: If f(x)f(x) and g(x)g(x) are differentiable functions, the derivative of their product, p(x)=f(x)⋅g(x)p(x) = f(x) \cdot g(x), is given by the formula: ddx(f(x)g(x))=f′(x)g(x)+f(x)g′(x)\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)

    • This can be remembered as 'first prime times second plus first times second prime'.

Quotient Rule

  • Definition: If f(x)f(x) and g(x)g(x) are differentiable functions and g(x)≠0g(x) \neq 0, the derivative of their quotient, q(x)=f(x)g(x)q(x) = \frac{f(x)}{g(x)}, is given by the formula: ddx(f(x)g(x))=f′(x)g(x)−f(x)g′(x)(g(x))2\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}

    • This can be remembered as 'low d-high minus high d-low all over low squared'.

Stationary Points

  • Definition: Stationary points in the domain of a function f(x)f(x) are the values of xx such that f′(x)=0f'(x) = 0.

  • Tangent Line at Stationary Points: At stationary points, the tangent line to the graph of f(x)f(x) is horizontal.

Finding Stationary Points - Examples

  • Example 1a: Find the stationary points of f(x)=(x2−3)exf(x) = (x^2 - 3)e^x.

    1. Find the derivative f′(x)f'(x) using the Product Rule:
      Let u=x2−3  ⟹  u′=2xu = x^2 - 3 \implies u' = 2x
      Let v=ex  ⟹  v′=exv = e^x \implies v' = e^x
      f′(x)=u′v+uv′=(2x)ex+(x2−3)exf'(x) = u'v + uv' = (2x)e^x + (x^2 - 3)e^x
      f′(x)=ex(2x+x2−3)=ex(x2+2x−3)f'(x) = e^x(2x + x^2 - 3) = e^x(x^2 + 2x - 3)

    2. Set f′(x)=0f'(x) = 0:
      ex(x2+2x−3)=0e^x(x^2 + 2x - 3) = 0
      Since exe^x is never zero, we must solve x2+2x−3=0x^2 + 2x - 3 = 0.

    3. Factor the quadratic equation:
      (x+3)(x−1)=0(x + 3)(x - 1) = 0

    4. Solve for xx:
      x=−3x = -3 or x=1x = 1

    • Stationary Points: The stationary points are at x=−3x = -3 and x=1x = 1.

  • Example 1b: Find the stationary points of y=2x−1x2+1y = \frac{2x - 1}{x^2 + 1}.

    1. Find the derivative y′y' using the Quotient Rule:
      Let f(x)=2x−1  ⟹  f′(x)=2f(x) = 2x - 1 \implies f'(x) = 2
      Let g(x)=x2+1  ⟹  g′(x)=2xg(x) = x^2 + 1 \implies g'(x) = 2x
      y′=f′(x)g(x)−f(x)g′(x)(g(x))2=2(x2+1)−(2x−1)(2x)(x2+1)2y' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} = \frac{2(x^2 + 1) - (2x - 1)(2x)}{(x^2 + 1)^2}
      y′=2x2+2−(4x2−2x)(x2+1)2=2x2+2−4x2+2x(x2+1)2y' = \frac{2x^2 + 2 - (4x^2 - 2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2}
      y′=−2x2+2x+2(x2+1)2=−2(x2−x−1)(x2+1)2y' = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} = \frac{-2(x^2 - x - 1)}{(x^2 + 1)^2}

    2. Set y′=0y' = 0:
      −2(x2−x−1)(x2+1)2=0\frac{-2(x^2 - x - 1)}{(x^2 + 1)^2} = 0
      This implies −2(x2−x−1)=0-2(x^2 - x - 1) = 0, or x2−x−1=0x^2 - x - 1 = 0.

    3. Solve for xx using the quadratic formula since it doesn't factor easily:
      x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
      Here, a=1,b=−1,c=−1a=1, b=-1, c=-1.
      x=−(−1)±(−1)2−4(1)(−1)2(1)=1±1+42=1±52x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}

    • Stationary Points: The stationary points are at x=1+52x = \frac{1 + \sqrt{5}}{2} and x=1−52x = \frac{1 - \sqrt{5}}{2}.

Applications of Product and Quotient Rules with Specific Values

Given that f(2)=2f(2) = 2, g(2)=3g(2) = 3, f′(2)=−1f'(2) = -1 and g′(2)=−4g'(2) = -4.

  • Example 2a: Instantaneous rate of change of p(x)=(x3−5x+1)g(x)p(x) = (x^3 - 5x + 1)g(x) at x=2x = 2.

    • Instantaneous rate of change is the derivative of the function at a specific point.

    1. Define components for Product Rule:
      Let u(x)=x3−5x+1  ⟹  u′(x)=3x2−5u(x) = x^3 - 5x + 1 \implies u'(x) = 3x^2 - 5
      Let v(x)=g(x)  ⟹  v′(x)=g′(x)v(x) = g(x) \implies v'(x) = g'(x)
      So, p(x)=u(x)v(x)p(x) = u(x)v(x).

    2. Find p′(x)p'(x) using the Product Rule:
      p′(x)=u′(x)v(x)+u(x)v′(x)=(3x2−5)g(x)+(x3−5x+1)g′(x)p'(x) = u'(x)v(x) + u(x)v'(x) = (3x^2 - 5)g(x) + (x^3 - 5x + 1)g'(x)

    3. Evaluate p′(2)p'(2):
      u(2)=23−5(2)+1=8−10+1=−1u(2) = 2^3 - 5(2) + 1 = 8 - 10 + 1 = -1
      u′(2)=3(2)2−5=3(4)−5=12−5=7u'(2) = 3(2)^2 - 5 = 3(4) - 5 = 12 - 5 = 7
      g(2)=3g(2) = 3 (given)
      g′(2)=−4g'(2) = -4 (given)
      p′(2)=(7)(3)+(−1)(−4)=21+4=25p'(2) = (7)(3) + (-1)(-4) = 21 + 4 = 25

    • Result: The instantaneous rate of change of p(x)p(x) at x=2x = 2 is 2525.

  • Example 2b: Slope of the tangent line to the graph of y=q(x)=f(x)g(x)+1y = q(x) = \frac{f(x)}{g(x)} + 1 when x=2x = 2.

    • The slope of the tangent line is the derivative of the function at a specific point.

    1. Find q′(x)q'(x). The derivative of a constant is 0.
      q′(x)=ddx(f(x)g(x))+ddx(1)q'(x) = \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) + \frac{d}{dx}(1)
      Use the Quotient Rule for the first term:
      q′(x)=f′(x)g(x)−f(x)g′(x)(g(x))2+0q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} + 0

    2. Evaluate q′(2)q'(2):
      f(2)=2f(2) = 2 (given)
      g(2)=3g(2) = 3 (given)
      f′(2)=−1f'(2) = -1 (given)
      g′(2)=−4g'(2) = -4 (given)
      q′(2)=(−1)(3)−(2)(−4)(3)2=−3−(−8)9=−3+89=59q'(2) = \frac{(-1)(3) - (2)(-4)}{(3)^2} = \frac{-3 - (-8)}{9} = \frac{-3 + 8}{9} = \frac{5}{9}

    • Result: The slope of the tangent line to q(x)q(x) at x=2x = 2 is 59\frac{5}{9}.

Higher Derivatives

  • Definition: Higher derivatives are derivatives of derivatives. They are denoted as follows:

    • First derivative: f′(t)=dfdtf'(t) = \frac{df}{dt} (rate of change)

    • Second derivative: f′′(t)=d2fdt2f''(t) = \frac{d^2f}{dt^2} (rate of change of the rate of change)

    • Third derivative: f′′′(t)=d3fdt3f'''(t) = \frac{d^3f}{dt^3}

    • Fourth derivative: f(4)(t)=d4fdt4f^{(4)}(t) = \frac{d^4f}{dt^4} (and so on for higher orders)

Example 1: Calculating Higher Derivatives

Let f(t)=t4−2et+2f(t) = t^4 - 2e^t + 2.

  • a. Find the following derivatives of f(t)f(t):

    • (i) First derivative, f′(t)f'(t):
      f′(t)=ddt(t4−2et+2)=4t3−2etf'(t) = \frac{d}{dt}(t^4 - 2e^t + 2) = 4t^3 - 2e^t

    • (ii) Second derivative, f′′(t)=d2fdt2f''(t) = \frac{d^2f}{dt^2}:
      f′′(t)=ddt(4t3−2et)=12t2−2etf''(t) = \frac{d}{dt}(4t^3 - 2e^t) = 12t^2 - 2e^t

    • (iii) Third derivative, f′′′(t)f'''(t):
      f′′′(t)=ddt(12t2−2et)=24t−2etf'''(t) = \frac{d}{dt}(12t^2 - 2e^t) = 24t - 2e^t

    • (iv) Fourth derivative, d4fdt4\frac{d^4f}{dt^4}:
      f(4)(t)=ddt(24t−2et)=24−2etf^{(4)}(t) = \frac{d}{dt}(24t - 2e^t) = 24 - 2e^t

  • b. Physical Meaning of f′(t)f'(t) and f′′(t)f''(t) for Position Function:

    • If f(t)f(t) represents the position of a particle moving on a straight line:

      • f′(t)f'(t) (velocity): This represents the instantaneous velocity of the particle at time tt. It indicates both the speed and the direction of motion.

      • f′′(t)f''(t) (acceleration): This represents the instantaneous acceleration of the particle at time tt. It indicates the rate at which the velocity of the particle is changing.

Derivatives of Trigonometric Functions

Definitions of Trigonometric Functions

  • Tangent: tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}

  • Cotangent: cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}

  • Secant: sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}

  • Cosecant: csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}

Proving Trigonometric Derivatives

Given that ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x and ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x. We will use the Quotient Rule where appropriate.

  • a. Prove ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x

    1. Write tan⁡x\tan x as a quotient: tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}.

    2. Apply the Quotient Rule with f(x)=sin⁡xf(x) = \sin x (so f′(x)=cos⁡xf'(x) = \cos x) and g(x)=cos⁡xg(x) = \cos x (so g′(x)=−sin⁡xg'(x) = -\sin x).
      ddx(tan⁡x)=(cos⁡x)(cos⁡x)−(sin⁡x)(−sin⁡x)(cos⁡x)2\frac{d}{dx}(\tan x) = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2}
      =cos⁡2x+sin⁡2xcos⁡2x= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}

    3. Use the Pythagorean Identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1:
      =1cos⁡2x= \frac{1}{\cos^2 x}

    4. Rewrite using the definition of secant:
      =(1cos⁡x)2=sec⁡2x= \left(\frac{1}{\cos x}\right)^2 = \sec^2 x

  • b. Prove ddx(cot⁡x)=−csc⁡2x\frac{d}{dx}(\cot x) = -\csc^2 x

    1. Write cot⁡x\cot x as a quotient: cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}.

    2. Apply the Quotient Rule with f(x)=cos⁡xf(x) = \cos x (so f′(x)=−sin⁡xf'(x) = -\sin x) and g(x)=sin⁡xg(x) = \sin x (so g′(x)=cos⁡xg'(x) = \cos x).
      ddx(cot⁡x)=(−sin⁡x)(sin⁡x)−(cos⁡x)(cos⁡x)(sin⁡x)2\frac{d}{dx}(\cot x) = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}
      =−sin⁡2x−cos⁡2xsin⁡2x= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}

    3. Factor out −1-1 and use the Pythagorean Identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1:
      =−(sin⁡2x+cos⁡2x)sin⁡2x=−1sin⁡2x= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x}

    4. Rewrite using the definition of cosecant:
      =−(1sin⁡x)2=−csc⁡2x= -\left(\frac{1}{\sin x}\right)^2 = -\csc^2 x

  • c. Prove ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x

    1. Write sec⁡x\sec x as a quotient: sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.

    2. Apply the Quotient Rule with f(x)=1f(x) = 1 (so f′(x)=0f'(x) = 0) and g(x)=cos⁡xg(x) = \cos x (so g′(x)=−sin⁡xg'(x) = -\sin x).
      ddx(sec⁡x)=(0)(cos⁡x)−(1)(−sin⁡x)(cos⁡x)2\frac{d}{dx}(\sec x) = \frac{(0)(\cos x) - (1)(-\sin x)}{(\cos x)^2}
      =0−(−sin⁡x)cos⁡2x=sin⁡xcos⁡2x= \frac{0 - (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x}

    3. Separate the terms to match the secant and tangent definitions:
      =sin⁡xcos⁡x⋅1cos⁡x= \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}
      =tan⁡x⋅sec⁡x= \tan x \cdot \sec x

    4. Rearrange for the standard form:
      =sec⁡xtan⁡x= \sec x \tan x

  • d. Prove ddx(csc⁡x)=−csc⁡xcot⁡x\frac{d}{dx}(\csc x) = -\csc x \cot x

    1. Write csc⁡x\csc x as a quotient: csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}.

    2. Apply the Quotient Rule with f(x)=1f(x) = 1 (so f′(x)=0f'(x) = 0) and g(x)=sin⁡xg(x) = \sin x (so g′(x)=cos⁡xg'(x) = \cos x).
      ddx(csc⁡x)=(0)(sin⁡x)−(1)(cos⁡x)(sin⁡x)2\frac{d}{dx}(\csc x) = \frac{(0)(\sin x) - (1)(\cos x)}{(\sin x)^2}
      =0−cos⁡xsin⁡2x=−cos⁡xsin⁡2x= \frac{0 - \cos x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x}

    3. Separate the terms to match the cosecant and cotangent definitions:
      =−cos⁡xsin⁡x⋅1sin⁡x= -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}
      =−cot⁡x⋅csc⁡x= -\cot x \cdot \csc x

    4. Rearrange for the standard form:
      =−csc⁡xcot⁡x= -\csc x \cot x

Application: Position, Velocity, and Acceleration with Trigonometric Functions

  • Problem: A piece of wood floating on a pond bobs according to s(t)=cos⁡(t)+sin⁡(t)s(t) = \cos(t) + \sin(t) cm, where tt is in seconds.

  • a. Find formulas for its velocity and acceleration at time tt seconds.

    • Position: s(t)=cos⁡(t)+sin⁡(t)s(t) = \cos(t) + \sin(t)

    • Velocity (v(t)=s′(t)v(t) = s'(t)): The derivative of position.
      v(t)=ddt(cos⁡(t)+sin⁡(t))=−sin⁡(t)+cos⁡(t)v(t) = \frac{d}{dt}(\cos(t) + \sin(t)) = -\sin(t) + \cos(t)

    • Acceleration (a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t)): The derivative of velocity (or second derivative of position).
      a(t)=ddt(−sin⁡(t)+cos⁡(t))=−cos⁡(t)−sin⁡(t)a(t) = \frac{d}{dt}(-\sin(t) + \cos(t)) = -\cos(t) - \sin(t)

  • b. Find the smallest time at which the velocity of the piece of wood is zero.

    1. Set the velocity function v(t)v(t) to zero:
      v(t)=−sin⁡(t)+cos⁡(t)=0v(t) = -\sin(t) + \cos(t) = 0

    2. Rearrange the equation:
      cos⁡(t)=sin⁡(t)\cos(t) = \sin(t)

    3. Divide by cos⁡(t)\cos(t) (assuming cos⁡(t)≠0\cos(t) \neq 0 to avoid undefined tangent):
      sin⁡(t)cos⁡(t)=1  ⟹  tan⁡(t)=1\frac{\sin(t)}{\cos(t)} = 1 \implies \tan(t) = 1

    4. Find the smallest positive value of tt for which tan⁡(t)=1\tan(t) = 1. This occurs in the first quadrant where both sine and cosine are positive and equal.
      t=π4t = \frac{\pi}{4}

    • Result: The smallest time at which the velocity is zero is t=π4t = \frac{\pi}{4} seconds.

Limits for Derivatives of Trigonometric Functions

  • Assumed Fundamental Limits:

    • lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

    • lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0

  • a. Find the values of:

    • (i) lim⁡x→0sin⁡7x3x\lim_{x \to 0} \frac{\sin 7x}{3x}

      1. To use the fundamental limit lim⁡u→0sin⁡uu=1\lim_{u \to 0} \frac{\sin u}{u} = 1, we need the argument of sine to match the denominator. Let u=7xu = 7x. Then as x→0x \to 0, u→0u \to 0.

      2. Manipulate the expression:
        lim⁡<em>x→0sin⁡7x3x=lim⁡</em>x→0(sin⁡7x7x⋅7x3x)(Multiply by 7x7x)\lim<em>{x \to 0} \frac{\sin 7x}{3x} = \lim</em>{x \to 0} \left(\frac{\sin 7x}{7x} \cdot \frac{7x}{3x}\right) \quad \text{(Multiply by }\frac{7x}{7x})
        =lim⁡<em>x→0(sin⁡7x7x)⋅lim⁡</em>x→0(73)= \lim<em>{x \to 0} \left(\frac{\sin 7x}{7x}\right) \cdot \lim</em>{x \to 0} \left(\frac{7}{3}\right)

      3. Apply the limit properties:
        =(1)⋅(73)=73= (1) \cdot \left(\frac{7}{3}\right) = \frac{7}{3}

    • (ii) lim⁡x→0tan⁡x2x\lim_{x \to 0} \frac{\tan x}{2x}

      1. Rewrite tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x}.
        lim⁡<em>x→0tan⁡x2x=lim⁡</em>x→0sin⁡x/cos⁡x2x\lim<em>{x \to 0} \frac{\tan x}{2x} = \lim</em>{x \to 0} \frac{\sin x / \cos x}{2x}
        =lim⁡x→0sin⁡x2xcos⁡x= \lim_{x \to 0} \frac{\sin x}{2x \cos x}

      2. Separate into known limits:
        =lim⁡<em>x→0(sin⁡xx⋅12cos⁡x)= \lim<em>{x \to 0} \left(\frac{\sin x}{x} \cdot \frac{1}{2 \cos x}\right) =lim⁡</em>x→0(sin⁡xx)⋅lim⁡x→0(12cos⁡x)= \lim</em>{x \to 0} \left(\frac{\sin x}{x}\right) \cdot \lim_{x \to 0} \left(\frac{1}{2 \cos x}\right)

      3. Apply the limit values (lim⁡x→0cos⁡x=cos⁡0=1\lim_{x \to 0} \cos x = \cos 0 = 1):
        =(1)⋅(12⋅1)=12= (1) \cdot \left(\frac{1}{2 \cdot 1}\right) = \frac{1}{2}

  • b. Show that the derivative of sin⁡x\sin x is cos⁡x\cos x.

    • Requirement: Use the identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B.

    1. Start with the limit definition of the derivative:
      f′(x)=lim⁡<em>h→0f(x+h)−f(x)hf'(x) = \lim<em>{h \to 0} \frac{f(x + h) - f(x)}{h} For f(x)=sin⁡xf(x) = \sin x: ddx(sin⁡x)=lim⁡</em>h→0sin⁡(x+h)−sin⁡xh\frac{d}{dx}(\sin x) = \lim</em>{h \to 0} \frac{\sin(x + h) - \sin x}{h}

    2. Apply the angle addition identity to sin⁡(x+h)\sin(x + h):
      ddx(sin⁡x)=lim⁡h→0(sin⁡xcos⁡h+cos⁡xsin⁡h)−sin⁡xh\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{(\sin x \cos h + \cos x \sin h) - \sin x}{h}

    3. Rearrange terms to group sin⁡x\sin x and cos⁡x\cos x:
      =lim⁡<em>h→0sin⁡xcos⁡h−sin⁡x+cos⁡xsin⁡hh= \lim<em>{h \to 0} \frac{\sin x \cos h - \sin x + \cos x \sin h}{h} =lim⁡</em>h→0sin⁡x(cos⁡h−1)+cos⁡xsin⁡hh= \lim</em>{h \to 0} \frac{\sin x (\cos h - 1) + \cos x \sin h}{h}

    4. Split the fraction into two separate limits:
      =lim⁡<em>h→0(sin⁡xcos⁡h−1h+cos⁡xsin⁡hh)= \lim<em>{h \to 0} \left( \sin x \frac{\cos h - 1}{h} + \cos x \frac{\sin h}{h} \right) =(sin⁡x)lim⁡</em>h→0cos⁡h−1h+(cos⁡x)lim⁡h→0sin⁡hh= (\sin x) \lim</em>{h \to 0} \frac{\cos h - 1}{h} + (\cos x) \lim_{h \to 0} \frac{\sin h}{h}

    5. Use the fundamental limits: Recall lim⁡<em>h→01−cos⁡hh=0\lim<em>{h \to 0} \frac{1 - \cos h}{h} = 0, so lim⁡</em>h→0cos⁡h−1h=0\lim</em>{h \to 0} \frac{\cos h - 1}{h} = 0. Also, lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1.
      =(sin⁡x)(0)+(cos⁡x)(1)= (\sin x)(0) + (\cos x)(1)
      =0+cos⁡x= 0 + \cos x
      =cos⁡x= \cos x

    • Conclusion: Therefore, ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x. This derivation relies on the foundational limit definitions for sine and cosine, and the trigonometric sum identity.