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Calculus AB : Basic Differentiation Rules

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Example Questions

Example Question #192 : Calculus Ab

Which of the following is the derivative of f(x) = sec(x) ?

Possible Answers:

tan^2(x)

cos(x)sin(x)

csc^2(x)

tan(x)sec(x)

Correct answer:

tan(x)sec(x)

Explanation:

Recall that $sec(x) = \frac{1}{cos(x)}$

f(x) = sec(x)

$f(x) = \frac{1}{cos(x)}$

$f(x) = cos(x)^{-1}$

$f'(x) = -cos(x)^{-2} * \frac{d}{dx}cos(x)$

$f'(x) = -\frac{1}{cos(x)^2} * -sin(x)$

$f'(x) = \frac{sin(x)}{cos(x)}*\frac{1}{cos(x)}$

f'(x) = tan(x)sec(x) $(tan(x) = \frac{sin(x)}{cos(x)}, sec(x) = \frac{1}{cos(x)})$

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Example Question #62 : Basic Differentiation Rules

Which of the following is the derivative of f(x) = csc(x) ?

Possible Answers:

tancsc(x)

sec^2(x)

cot^2(x)

-cot(x)csc(x)

Correct answer:

-cot(x)csc(x)

Explanation:

Recall that $csc(x) = \frac{1}{sin(x)}$

f(x) = csc(x)

$f(x) = \frac{1}{sin(x)}$

$f(x) = sin(x)^{-1}$

$f'(x) = -sin(x)^{-2}*\frac{d}{dx}sin(x)$

$f'(x) = -\frac{1}{sin(x)^2}*cos(x)$

$f'(x) = -\frac{cos(x)}{sin(x)}*\frac{1}{sin(x)}$

f'(x) = -cot(x)csc(x) $(\csc(x) = \frac{1}{sin(x)}, cot(x) = \frac{cos(x)}{sin(x)})$

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Example Question #191 : Calculus Ab

What is the derivative for f(x) = cot(x) ?

Possible Answers:

csc^2(x)

sec^2(x)

cot^2(x)

tan^2(x)

Correct answer:

csc^2(x)

Explanation:

Recall that $cot(x) = \frac{1}{tan(x)} = \frac{cos(x)}{sin(x)}$ . We are able to use the quotient rule to find the derivative now.

f(x) = cot(x)

$f(x) = \frac{cos(x)}{sin(x)}$

$f'(x) = \frac{-sin(x)sin(x) - cos(x)cos(x)}{sin^2(x)}$

$f'(x) = \frac{-sin^2(x) - cos^2(x)}{sin^2(x)}$

$f'(x) = \frac{-(sin^2(x) - cos^2(x))}{sin^2(x)}$

$f'(x) = \frac{-1}{sin^2(x)}$ (Pythagorean Identity)

f'(x) = csc^2(x) $(csc = \frac{1}{sin(x)})$

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Example Question #193 : Calculus Ab

True or False: We use the Chain Rule to find the derivatives of trigonometric functions.

Possible Answers:

True

False

Correct answer:

True

Explanation:

This is true. The chain rule is what allows us to differentiate composite functions. Composite functions are functions that are within one another. Take tan(2x) for example. tan(x) in itself is a function, but so is 2(x) . We can let f(x) = tan(x) and g(x) = 2x . So the composite function would be, f(g(x)) = tan(2x) .

Now for the Chain Rule we multiply the derivative of the inside function by the coefficient of the outside function while differentiating the outside function as well. F'(x) will be the composite function f(g(x)) . The formula for the Chain Rule is:

F'(x) = g'(x)f'(g(x)) .

Going back to our example of tan(2x) , we know that the derivative of tan(x) is sec^2(x) and the derivative of is . So the derivative of our composite function is 2sec^2(2x) . Thus, every time we are differentiating trigonometric functions we are utilizing the Chain Rule whether we realize it or not.

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Example Question #2 : Use Derivatives Of Trig Functions

What is the derivative of f(x) = tan(x^2 + 3x) ?

Possible Answers:

f'(x) = sec^2(x^2 + 3x)

f'(x) = cot^2(2x + 3)

f'(x) = sec^2(2x + 3)

f'(x) = (2x + 3)sec^2(x^2 + 3x)

Correct answer:

f'(x) = (2x + 3)sec^2(x^2 + 3x)

Explanation:

We must take the derivative of both the quantity of the tangent function and tangent itself to solve for this derivative. Recall that the derivative of tan(x) = sec^2(x)

f(x) = tan(x^2 + 3x)

$f'(x) = \frac{d}{dx}[x^2+3x]*\frac{d}{dx}[tan(x^2 + 3x)]$

f'(x) = (2x + 3)sec^2(x^2 + 3x)

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Example Question #192 : Calculus Ab

What is the derivative of f(x) = 2x + csc(4x) at $x = \pi$ ?

Possible Answers:

7.25

This derivative is undefined at this point

$\pi$

Correct answer:

This derivative is undefined at this point

Explanation:

We must start this problem by finding the derivative. Recall that the derivative of the cosecant function is csc(x) = -cot(x)csc(x)

f(x) = 2x + csc(4x)

$f'(x) = 2 + 4*\frac{d}{dx}[csc(4x)]$

f'(x) = 2 - 4cot(4x)csc(4x)

Now we must find the derivative of $x = \pi$ . So we will plug $\pi$ in for .

$f'(\pi) = 2 - 4cot(4\pi)csc(4\pi)$

$f'(\pi) = 2 - 4(0)(undefined)$

$f'(\pi) = undefined$

And so at the point of the graph of the derivative, the function is undefined.

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Example Question #61 : Basic Differentiation Rules

True or False: When the derivative of tan(x) is negative but increasing, then the function tan(x) is increasing.

Possible Answers:

True

False

Correct answer:

False

Explanation:

Even though we are working with trigonometric functions, the rules of derivatives remain the same. Since the graph of a derivative function is the graph of the rate of change of the original function, then when the graph of the derivative function is negative, the original function is decreasing. Even when the derivative is increasing, if it is negative the original function is still decreasing, just by less of a factor. When the derivative function is at zero, then the original function is constant. When the derivative function is positive, then the original function is increasing.

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Example Question #2 : Use Derivatives Of Trig Functions

Find the derivative of f(x) = cot(5x^2) .

Possible Answers:

f'(x) = 5x^2csc^2(5x^2)

f'(x) = 5csc^2(5x^2)

f'(x) = csc^2(5x^2)

f'(x) = 10xcsc^2(5x^2)

Correct answer:

f'(x) = 10xcsc^2(5x^2)

Explanation:

We need to use the Chain Rule to take both the derivative of the trigonometric function and the quantity within the trig function.

f(x) = cot(5x^2)

$f'(x) = csc^2(5x^2) * \frac{d}{dx}[5x^2]$

f'(x) = 10xcsc^2(5x^2)

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Example Question #2 : Use Derivatives Of Trig Functions

What is the derivative of f(x) = tan^2(x) ?

Possible Answers:

f'(x) = sec^2(x)tan(x)

f'(x) = 2tan(x)

f'(x) = 2sec^2(x)tan(x)

f'(x) = 2sec^2(x)

Correct answer:

f'(x) = 2sec^2(x)tan(x)

Explanation:

Recall that the derivative of the tangent function is sec(x) . So we need to take the derivative of the exponent of tan^2(x) and we also need the derivative of the actual tangent function.

f(x) = tan^2(x)

$f'(x) = 2tan(x) *\frac{d}{dx}[tan(x)]$

f'(x) = 2sec^2(x)tan(x)

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Example Question #191 : Calculus Ab

Which of the following is the derivative of f(x) = tan(x) ?

Possible Answers:

cot^2(x)

sec^2(x)

csc^2(x)

cos^2(x)

Correct answer:

sec^2(x)

Explanation:

Recall that the function $tan(x) = \frac{sin(x)}{cos(x)}$ . Knowing this, we can use the quotient rule to find the derivative.

f(x) = tan(x)

$f(x) = \frac{sin(x)}{cos(x)}$

$f'(x) = \frac{cos(x)cos(x) - -sin(x)sin(x)}{cos^2(x)}$

$f'(x) = \frac{cos^2(x) + sin^2(x)}{cos^2(x)}$

$f'(x) = \frac{1}{cos^2(x)}$ (Pythagorean Identity)

f'(x) = sec^2(x) $(\frac{1}{cos(x)} = sec(x))$

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Calculus AB : Basic Differentiation Rules

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #192 : Calculus Ab

Example Question #62 : Basic Differentiation Rules

Example Question #191 : Calculus Ab

Example Question #193 : Calculus Ab

Example Question #2 : Use Derivatives Of Trig Functions

Example Question #192 : Calculus Ab

Example Question #61 : Basic Differentiation Rules

Example Question #2 : Use Derivatives Of Trig Functions

Example Question #2 : Use Derivatives Of Trig Functions

Example Question #191 : Calculus Ab

All Calculus AB Resources

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