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Calculus 1 : Calculus

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Example Questions

Example Question #202 : Differential Functions

What is the first derivative of f(x) = x²sin(4x³)?

Possible Answers:

2x(sin(4x³) + 6x³cos(4x³))

sin(4x³) + 12x²cos(4x³)

2x * cos(4x³)

24x⁴ * cos(4x³)

2x * sin(4x³) + 12x⁵cos(4x³)

Correct answer:

2x(sin(4x³) + 6x³cos(4x³))

Explanation:

This is a product rule combined with a chain rule. Let's do the chain rule for sin(4x³) first:

cos(4x³) * 12x² = 12x²cos(4x³)

With this in mind, let's solve our whole problem:

2x * sin(4x³) + x² * 12x²cos(4x³) = 2x * sin(4x³) + 12x⁴cos(4x³) = 2x(sin(4x³) + 6x³cos(4x³))

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Example Question #203 : Differential Functions

What is the first derivative of f(x) = x⁴ – x * sin(x^–5)?

Possible Answers:

4x³ – cos(x^–5)

4x³ + cos(x^–5)

4x³ + sin(x^–5) – (5 * cos(x^–5)/x⁵)

4x³ – sin(x^–5) + (5 * cos(x^–5)/x⁵)

None of the other answers

Correct answer:

4x³ – sin(x^–5) + (5 * cos(x^–5)/x⁵)

Explanation:

The first element is merely differentiated as 4x³

The second element is a relatively simple product rule:

sin(x^–5) + x * cos(x^–5) * –5 * x^–6 = sin(x^–5) + cos(x^–5) * –5 * x^–5 = sin(x^–5) – (5 * cos(x^–5)/x⁵)

Put everything back together:

4x³ – ( sin(x^–5) – (5 * cos(x^–5)/x⁵) ) = 4x³ – sin(x^–5) + (5 * cos(x^–5)/x⁵)

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Example Question #204 : Differential Functions

What is the first derivative of f(x) = 5x * ln(2x)?

Possible Answers:

5ln(2)/2x

5/x

5(ln(2) + (1/2))

5/2x

5(ln(2) + 1)

Correct answer:

5(ln(2) + 1)

Explanation:

This is just a normal product rule problem:

5 * ln(2x) + 5x * 2 * (1/2x)

Simplify: 5ln(2) + 5 = 5(ln(2) + 1)

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Example Question #204 : Differential Functions

What is the slope of the tangent line at x = 2 for f(x) = 6/x²?

Possible Answers:

1.5

–12

–3

–1.5

Correct answer:

–1.5

Explanation:

First rewrite your function to make this easier:

f(x) = 6/x²= 6x^–2

Now, we must find the first derivative:

f'(x) = –2 * 6 * x^–3 = –12/x³

The slope of the tangent line of f(x) at x = 2 is: f'(2) = –12/2³ = –12/8 = –3/2 = –1.5

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Example Question #205 : Differential Functions

What is the first derivative of f(x) = 2ln(cos(x)sin(x))?

Possible Answers:

2cos²(x)sec(x)csc(x)

None of the other answers

2cot(x)cos²(x)

cot(x)cos²(x)

2cos(2x)sec(x)csc(x)

Correct answer:

2cos(2x)sec(x)csc(x)

Explanation:

This requires both the use of the chain rule and the product rule. Start with the natural logarithm: 2/(cos(x)sin(x))

Now, multiply by d/dx cos(x)sin(x), which is: –sin(x)sin(x) + cos(x)cos(x) = cos²(x) – sin²(x)

Therefore, f'(x) = (cos²(x) – sin²(x)) * 2/(cos(x)sin(x)) = 2(cos²(x) – sin²(x))sec(x)csc(x)

From our trigonometric identities, we know cos²(x) – sin²(x) = cos(2x)

Therefore, we can finilize our simplification to f'(x) = 2cos(2x)sec(x)csc(x)

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Example Question #14 : How To Find Differential Functions

Take the derivative of

$y=(7+x)^{(7+x)^2}$

Possible Answers:

${y}'=(7+x)^{(7+x)^2}(1+\ln(7+x)^{(7+x)^2)})$

${y}'=(7+x)^{(7+x)^2}[21+\ln(7+x)+x]$

Derivative does not exist

${y}'=(7+x)^2(7+x)^{(7+x)^2-1}$

${y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]$

Correct answer:

${y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]$

Explanation:

We need to use logarithm differentiation to do this problem. Take the natural log of both sides.

$\ln y=\ln(7+x)^{(7+x)^2}$

Apply the power rule of natural log

$\ln y=(7+x)^2\ln(7+x)$

Perform implicit differentiation to both sides

$\frac{{y}'}{y}=2(7+x)\ln(7+x)+7+x$

Solve for $y^{'}$

${y}'=y[2(7+x)\ln(7+x)+7+x]$

${y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]$

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Example Question #21 : How To Find Differential Functions

Find $\small d(\tan x)/dx$

Possible Answers:

$\small \sec^2x$

$\small -\sec^2x$

$\small \csc^2x$

$\small -\csc^2x$

$\small \cot^2x$

Correct answer:

$\small \sec^2x$

Explanation:

$\small \frac{d}{dx}(\tan x) = \frac{d}{dx}(\frac{\sin x}{\cos x}) = \frac {\cos x \frac{d}{dx}(\sin x) - \sin x \frac{d}{dx}(\cos x)}{\cos^2x}$

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

$\small = \frac{\cos^2x + \sin^2x}{\cos^2x}$
$\small = \frac{1}{\cos^2x} = \sec^2x$

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Example Question #208 : Differential Functions

Differentiate:

$\small \sin(x^2+x)$ with respect to $\small x$ .

Possible Answers:

$\small (2x+1)(\cos(x^2+x))$

$\small -(2x+1)(cos(x^2+x))$

$\small (x^2+x)(\sin(2x+1))$

$\small (x^2+x)(\cos(2x+1))$

$\small (2x+1)(\sin(x^2+x))$

Correct answer:

$\small (2x+1)(\cos(x^2+x))$

Explanation:

Apply the chain rule: differentiate the "outside" function first. Let $u=x^{2}+x$ .
$\small \small \small \small \frac{d}{du} \sin(u) = \cos(u) = \cos(x^2+x)$
Differentiate the "inside" function next.
$\small \frac{d}{dx}(x^2+x) = 2x+1$
Multiply these two functions to find the derivative of the original function.
$\small \cos(x^2+x) * (2x+1)$

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Example Question #203 : Functions

Evaluate $\small \int (x^2 - 3x + 6) dx$

Possible Answers:

$\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

$\small \small \small x^3 - 3x^2 + 6x + C$

$\small 2x-3+C$

$\small \small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x$

$\small 2x - 3$

Correct answer:

$\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

Explanation:

To integrate the function, integrate each term of the function. e.g., integrate $\small x^2$ by increasing the exponent by 1 integer and dividing the term by this new integer: $\small \frac {x^3}{3}$ .

Do this for the rest to get $\small \small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x$ .

But remember that every integration requires an arbitrary constant, $\small C$ . Thus, the integral of the function is $\small \frac{x^3}{3} - \frac{3x^2}{2}+ 6x + C$

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Example Question #211 : Differential Functions

Integrate

$\small \int\sin^2x\hspace2dx$

Possible Answers:

$\small \frac{x}{2}-\frac{\sin2x}{4}+C$

$\small \small \small \frac{x}{2}+\frac{\sin2x}{4}$

$\small \small \small \frac{x}{2}+\frac{\cos2x}{4}+C$

$\small \small \small \frac{x}{2}-\frac{\cos2x}{4}+C$

$\small \small \frac{x}{2}+\frac{\sin2x}{4}+C$

Correct answer:

$\small \frac{x}{2}-\frac{\sin2x}{4}+C$

Explanation:

We can use trigonometric identities to transform integrals that we typically don't know how to integrate.

$\small \small \sin^2x = \frac{1-\cos2x}{2}$
Thus,
$\small \small \small \int\sin^2x\hspace2dx = \int \frac{1-\cos2x}{2}\hspace2dx$
$\small = \frac{1}{2} \int (1-\cos2x) \hspace2 dx = \frac{1}{2}x-\frac{1}{2}\frac{sin2x}{2}+C = \frac{x}{2} - \frac{sin2x}{4}+C$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #202 : Differential Functions

Example Question #203 : Differential Functions

Example Question #204 : Differential Functions

Example Question #204 : Differential Functions

Example Question #205 : Differential Functions

Example Question #14 : How To Find Differential Functions

Example Question #21 : How To Find Differential Functions

Example Question #208 : Differential Functions

Example Question #203 : Functions

Example Question #211 : Differential Functions

All Calculus 1 Resources

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