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Grade 12 · MCV4U · Calculus & Vectors

MCV4U Derivatives of Exponential and Trig Functions Practice Test (with Solutions)

This unit adds e^x, b^x, ln x, sine and cosine to the functions you can differentiate. Expect chain, product and quotient rule questions built from them, exact exponential and log equations, tangent lines with a condition on the slope, and a proof or two. Try each question below on paper before you open the answer.

Free Derivatives of Trig, Exp and Log Practice Test

8 questions (15 parts), exactly as a free account gets them in the first Derivatives of Trig, Exp and Log test on Martian Lab. Try each one on paper first, then open the answer.

  1. Is the derivative of a sinusoidal function always periodic? Explain why or why not with an example.
    Show solution
    The derivative of sin⁡(x)\sin(x) is cos⁡(x)\cos(x) and the derivative of cos⁡(x)\cos(x) is −sin⁡(x)-\sin(x). Both derivative functions are periodic. Since all sinusoidal functions can be expressed as sin⁡(x)\sin(x) or cos⁡(x)\cos(x), their derivative would also be periodic. y=tan⁡xy = \tan x, y′=sec⁡2xy' = \sec^2 x is also periodic.
  2. Let f(x)=bxf(x) = b^x, where b>1b > 1, and let g(x)=f′(x)f(x)g(x) = \dfrac{f'(x)}{f(x)}.
    1. Predict the equation and the type of graph of g(x)g(x) by giving one example.
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      It would be a constant. f(x)=2xf(x) = 2^x, f′(x)=2x⋅ln⁡2f'(x) = 2^x \cdot \ln 2 g(x)=2x⋅ln⁡22x=ln⁡2g(x) = \dfrac{2^x \cdot \ln 2}{2^x} = \ln 2
    2. For what value of bb will g(x)=eg(x) = e?
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      f(x)=bxf(x) = b^x f′(x)=bx⋅ln⁡bf'(x) = b^x \cdot \ln b f′(x)f(x)=ln⁡b=e\dfrac{f'(x)}{f(x)} = \ln b = e, b=eeb = e^e.
  3. Solve for xx and state your solutions in exact value.
    1. e−3x+8=12e^{-3x+8} = 12
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      −3x+8=ln⁡12-3x+8 = \ln 12 3x=8−ln⁡123x = 8 - \ln 12 x=8−ln⁡123x = \dfrac{8 - \ln 12}{3}
    2. ln⁡(ln⁡(7x−5))=−4\ln(\ln(7x-5)) = -4
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      ln⁡(7x−5)=e−4\ln(7x-5) = e^{-4} 7x−5=ee−47x-5 = e^{e^{-4}} x=ee−4+57x = \dfrac{e^{e^{-4}}+5}{7}
    3. ln⁡(3x−2)=−5\ln(3x-2) = -5
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      3x−2=e−53x-2 = e^{-5} x=e−5+23x = \dfrac{e^{-5}+2}{3}
  4. Differentiate but DO NOT simplify.
    1. y=sin⁡ ⁣(π3−2x)y = \sin\!\left(\dfrac{\pi}{3} - 2^x\right)
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      y′=cos⁡ ⁣(π3−2x)⋅(−2x)⋅ln⁡2y' = \cos\!\left(\dfrac{\pi}{3} - 2^x\right) \cdot \left(-2^x\right) \cdot \ln 2
    2. y=π22sin⁡ ⁣(xπ)y = \dfrac{\pi^2}{2}\sin\!\left(\dfrac{x}{\pi}\right)
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      y′=π22⋅cos⁡ ⁣(xπ)⋅1πy' = \dfrac{\pi^2}{2} \cdot \cos\!\left(\dfrac{x}{\pi}\right) \cdot \dfrac{1}{\pi}
    3. y=e2x+sin⁡xy = \sqrt{e^{2x} + \sin x}
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      y′=12 ⁣(e2x+sin⁡x)−12⋅(e2x⋅2+cos⁡x)y' = \dfrac{1}{2}\!\left(e^{2x}+\sin x\right)^{-\frac{1}{2}} \cdot \left(e^{2x} \cdot 2 + \cos x\right)
    4. y=xsin⁡(4x)y = x\sin(4x)
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      y′=sin⁡(4x)+x⋅cos⁡(4x)⋅4y' = \sin(4x) + x \cdot \cos(4x) \cdot 4
    5. Use the Quotient Rule y=2xsin⁡2xy = \dfrac{2^x}{\sin^2 x}
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      y′=2xln⁡2⋅sin⁡2x−2x⋅2sin⁡x⋅cos⁡xsin⁡4xy' = \dfrac{2^x \ln 2 \cdot \sin^2 x - 2^x \cdot 2\sin x \cdot \cos x}{\sin^4 x}
  5. If f(x)=13e6x+4sin⁡(x)f(x) = \dfrac{1}{3}e^{6x} + 4\sin(x), find f′′ ⁣(π3)f''\!\left(\dfrac{\pi}{3}\right).
    Show solution
    f′(x)=13e6x⋅6+4cos⁡(x)=2⋅e6x+4cos⁡(x)f'(x) = \dfrac{1}{3}e^{6x} \cdot 6 + 4\cos(x) = 2 \cdot e^{6x} + 4\cos(x) f′′(x)=2e6x⋅6−4sin⁡(x)=12e6x−4sin⁡(x)f''(x) = 2e^{6x} \cdot 6 - 4\sin(x) = 12e^{6x} - 4\sin(x) f′′ ⁣(π3)=12⋅e2π−4⋅sin⁡ ⁣(π3)=12e2π−23f''\!\left(\dfrac{\pi}{3}\right) = 12 \cdot e^{2\pi} - 4 \cdot \sin\!\left(\dfrac{\pi}{3}\right) = 12e^{2\pi} - 2\sqrt{3}.
  6. Given f(x)=−2sin⁡(3x)f(x) = -2\sin(3x), determine the equation of the tangent line to the graph of f(x)f(x) with Use EXACT radian measure where slope is minimum.
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    f′(x)=−2cos⁡(3x)⋅3=−6cos⁡(3x)f'(x) = -2\cos(3x) \cdot 3 = -6\cos(3x) minimum slope is −6-6, it occurs at x=0x = 0. x=0x = 0, y=0y = 0 ∴\therefore The eq. of tangent is y=−6xy = -6x.
  7. Find the equation of the tangent line to the graph of f(x)=e−x2f(x) = e^{-\frac{x}{2}} that is perpendicular to the line x−2y−6=0x - 2y - 6 = 0. Express your answer in EXACT value.
    Show solution
    2y=x−62y = x - 6, y=12x−3y = \dfrac{1}{2}x - 3 ∴  m⊥=−2\therefore\; m_\perp = -2 f′(x)=e−x2⋅12=−2f'(x) = e^{-\frac{x}{2}} \cdot \dfrac{1}{2} = -2 e−x2=4e^{-\frac{x}{2}} = 4 −x2=ln⁡4\dfrac{-x}{2} = \ln 4 x=−2ln⁡4x = -2\ln 4 y=e−12(−2ln⁡4)=4y = e^{-\frac{1}{2}(-2\ln 4)} = 4 y=mx+by = mx + b 4=(−2)(−2ln⁡4)+b4 = (-2)(-2\ln 4) + b b=4−4ln⁡4b = 4 - 4\ln 4 ∴\therefore The equation of tangent is y=−2x+(4−4ln⁡4)y = -2x + (4 - 4\ln 4).
  8. Show that f(x)=1−cos⁡(x)sin⁡(x)f(x) = \dfrac{1-\cos(x)}{\sin(x)} has no horizontal tangent within its domain.
    Show solution
    f′(x)=sin⁡x⋅sin⁡x−(1−cos⁡x)⋅cos⁡xsin⁡2xf'(x) = \dfrac{\sin x \cdot \sin x - (1-\cos x)\cdot \cos x}{\sin^2 x} =sin⁡2x+cos⁡2x−cos⁡xsin⁡2x= \dfrac{\sin^2 x + \cos^2 x - \cos x}{\sin^2 x} =1−cos⁡xsin⁡2x= \dfrac{1 - \cos x}{\sin^2 x} For f′(x)=0f'(x) = 0, cos⁡x=1\cos x = 1. However, when cos⁡x=1\cos x = 1, sin⁡x=0\sin x = 0. It is not within the domain. ∴\therefore There is no horizontal tangent.

This Unit on Martian Lab

The lessons cover the derivatives of logarithmic and exponential functions and their applications, then sinusoidal functions and their applications, and finish with mixed functions.

  1. Derivatives of Logarithmic & Exponential Functions
  2. Applications of Logarithmic & Exponential Functions
  3. Derivatives of Sinusoidal Functions
  4. Applications of Sinusoidal Functions
  5. Derivatives of Mixed Functions

Then 6 full Derivatives of Trig, Exp and Log practice tests to check yourself.

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Derivatives of Trig, Exp and Log FAQ

What is on an MCV4U derivatives of exponential and trig functions test?+

Differentiating exponential, logarithmic, sine and cosine functions, usually combined with the chain, product and quotient rules. Also solving exponential and log equations exactly, second derivatives, tangent lines with a given slope, and showing a graph has no horizontal tangent. The questions on this page cover all of these.

What is the derivative of b^x?+

b^x ln b. For e^x that is just e^x, because ln e = 1. With the chain rule, the derivative of b^(g(x)) is b^(g(x)) ln b · g'(x).

What are the derivatives of sin x and cos x?+

The derivative of sin x is cos x and the derivative of cos x is -sin x. With an inner function, multiply by its derivative: the derivative of sin(4x) is 4 cos(4x), so x sin(4x) differentiates to sin(4x) + 4x cos(4x).

What is the derivative of ln x?+

1/x, for x > 0. With the chain rule, the derivative of ln(g(x)) is g'(x)/g(x).

Is the derivative of a sinusoidal function periodic?+

Yes, with the same period. The derivative of a sin(kx) + c is ak cos(kx), which repeats every 2π/k, just like the original.

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