MTH 102 Test Compilation 8

Here's a comprehensive study summary covering the calculus questions from the provided document, complete with solutions and explanations.

Study Summary: Calculus Questions

1. Differentiation Questions

1.1. Derivative of xtan x

Question: Find `dy/dx` if `y = x^(tan x)` (Appears as Question 6, 10, 2, 5 in different pages).

Solution: This requires logarithmic differentiation. Let `y = x^(tan x)`. Take the natural logarithm of both sides: `ln y = ln(x^(tan x))` Using logarithm property `ln(a^b) = b ln a`: `ln y = tan x * ln x`

Now, differentiate both sides with respect to `x` using the chain rule on the left and the product rule on the right:

`(1/y) * dy/dx = (d/dx(tan x)) * ln x + tan x * (d/dx(ln x))`

Recall: `d/dx(tan x) = sec^2 x` and `d/dx(ln x) = 1/x`

`(1/y) * dy/dx = sec^2 x * ln x + tan x * (1/x)`

`dy/dx = y * (sec^2 x * ln x + (tan x)/x)`

Substitute `y = x^(tan x)` back:

`dy/dx = x^(tan x) * (sec^2 x * ln x + (tan x)/x)`

Rearranging the terms to match the options:

`dy/dx = x^(tan x) * ((tan x)/x + sec^2 x * ln x)`

Correct Option: Option d. `dy/dx = x^(tan x) * (tan x / x + sec^2 x * ln x)`

1.2. Derivative of e^(2x) log(x+6)

Question: Find `dy/dx` if `y = e^(2x) log(x+6)` (Appears as Question 1, 9, 5, 4, 3, 7 in different pages).

Solution: This requires the product rule: `d/dx(u*v) = u'v + uv'` Let `u = e^(2x)` and `v = log(x+6)` (assuming `log` means natural logarithm `ln`).

Find `u'` and `v'`:

• `u' = d/dx(e^(2x))` = `e^(2x) * d/dx(2x)` = `2e^(2x)`
• `v' = d/dx(ln(x+6))` = `1/(x+6) * d/dx(x+6)` = `1/(x+6)`

Apply the product rule:

`dy/dx = u'v + uv'` `dy/dx = 2e^(2x) * ln(x+6) + e^(2x) * (1/(x+6))`

Factor out `e^(2x)`:

`dy/dx = e^(2x) * [2 ln(x+6) + 1/(x+6)]`

Or, written as in the options:

`dy/dx = e^(2x) * [1/(x+6) + 2 log(x+6)]`

Correct Option: Option a. `dy/dx = e^(2x) * [1/(x+6) + 2 log(x+6)]`

1.3. Derivative of ln(x^2 / (x^2 + 1))

Question: Obtain `d/dx {ln(x^2 / (x^2 + 1))}` (Appears as Question 2, 10 in different pages).

Solution: First, simplify the logarithm using properties: `ln(a/b) = ln a - ln b` and `ln(a^b) = b ln a`.

`ln(x^2 / (x^2 + 1)) = ln(x^2) - ln(x^2 + 1)` `= 2 ln x - ln(x^2 + 1)`

Now, differentiate with respect to `x`:

`d/dx [2 ln x - ln(x^2 + 1)]` `= 2 * (1/x) - (1/(x^2 + 1)) * d/dx(x^2 + 1)` `= 2/x - (1/(x^2 + 1)) * (2x)` `= 2/x - (2x)/(x^2 + 1)`

Combine the terms by finding a common denominator:

`= [2(x^2 + 1) - 2x(x)] / [x(x^2 + 1)]` `= [2x^2 + 2 - 2x^2] / [x(x^2 + 1)]` `= 2 / [x(x^2 + 1)]`

Correct Option: Option a. `2 / (x(x^2+1))`

1.4. Implicit Differentiation: 3x^2y + xy^2 = 5

Question: Find `dy/dx` if `3x^2y + xy^2 = 5` (Appears as Question 7, 2 in different pages).

Solution: Differentiate both sides with respect to `x`. Remember that `y` is a function of `x`, so use the product rule and chain rule where appropriate.

`d/dx(3x^2y) + d/dx(xy^2) = d/dx(5)`

For `d/dx(3x^2y)` (product rule: `u=3x^2`, `v=y`): `3 * (2x * y + x^2 * dy/dx) = 6xy + 3x^2 dy/dx`

For `d/dx(xy^2)` (product rule: `u=x`, `v=y^2`): `1 * y^2 + x * (2y * dy/dx) = y^2 + 2xy dy/dx`

For `d/dx(5)`: `0`

Combine the differentiated terms:

`6xy + 3x^2 dy/dx + y^2 + 2xy dy/dx = 0`

Group terms with `dy/dx` and terms without it:

`(3x^2 + 2xy) dy/dx = -6xy - y^2`

Factor out `-1` from the right side:

`(3x^2 + 2xy) dy/dx = -(y^2 + 6xy)`

Solve for `dy/dx`:

`dy/dx = -(y^2 + 6xy) / (3x^2 + 2xy)`

Correct Option: Option b. `dy/dx = -(y^2 + 6xy) / (3x^2 + 2xy)`

1.5. Implicit Differentiation: x^2 sin y - e^(2y) cos x = 3x^2 + 2x - 5

Question: Differentiate the implicit functions `x^2 sin y - e^(2y) cos x = 3x^2 + 2x - 5` with respect to `x` (Appears as Question 8, 7 in different pages).

Solution: Differentiate each term with respect to `x`.

1. `d/dx(x^2 sin y)`: (Product rule: `u=x^2`, `v=sin y`) `2x sin y + x^2 cos y * dy/dx`

2. `d/dx(-e^(2y) cos x)`: (Product rule: `u=-e^(2y)`, `v=cos x`) `- ( (e^(2y) * 2 dy/dx) * cos x + e^(2y) * (-sin x) )` `= -2e^(2y) cos x dy/dx + e^(2y) sin x`

3. `d/dx(3x^2 + 2x - 5)`: `6x + 2`

Combine the differentiated terms:

`2x sin y + x^2 cos y dy/dx - 2e^(2y) cos x dy/dx + e^(2y) sin x = 6x + 2`

Group terms with `dy/dx` and terms without it:

`(x^2 cos y - 2e^(2y) cos x) dy/dx = 6x + 2 - 2x sin y - e^(2y) sin x`

Solve for `dy/dx`:

`dy/dx = (6x + 2 - 2x sin y - e^(2y) sin x) / (x^2 cos y - 2e^(2y) cos x)`

Factor out `-1` from the numerator to rearrange terms for matching options:

`dy/dx = -(2x sin y + e^(2y) sin x - 6x - 2) / (x^2 cos y - 2e^(2y) cos x)`

The options provided have a different structure. Let's re-examine the options and the derived form. The general form of the correct answer should be: `dy/dx = (6x + 2 - 2x sin y - e^(2y) sin x) / (x^2 cos y - 2e^(2y) cos x)`

Let's check option `a` from one of the images: `dy/dx = (6x - 2x sin y - sin x e^(2y) + 2) / (x^2 cos y + 2 cos x e^(2y))`. This seems to have a sign error in the denominator. Let's check my differentiation of `-e^(2y)cos x`. `d/dx(-e^(2y)cos x) = -[(d/dx(e^(2y)))cos x + e^(2y)(d/dx(cos x))]` `= -[ (2e^(2y) dy/dx)cos x + e^(2y)(-sin x) ]` `= -2e^(2y) cos x dy/dx + e^(2y) sin x` This matches. So the denominator is `x^2 cos y - 2e^(2y) cos x`. And the numerator is `6x + 2 - 2x sin y - e^(2y) sin x`. Comparing to option `a`: `(6x - 2x sin y - sin x e^(2y) + 2) / (x^2 cos y + 2 cos x e^(2y))` The numerator matches after reordering. The denominator `x^2 cos y + 2 cos x e^(2y)` has a `+` instead of `-`. This suggests a potential sign error in the question's options or a misunderstanding of one of the terms' differentiation. However, given the structure and common patterns in such questions, let's assume one of the options is correct. If the question implies `d/dy` for the second term, or if `y` is some specific function of `x` such that `cos x` changes sign, it might be different. Let's re-verify: `d/dx(e^(2y) cos x)`: `(2e^(2y) dy/dx) cos x + e^(2y) (-sin x)` So, `d/dx(-e^(2y) cos x)` is `-2e^(2y) cos x dy/dx + e^(2y) sin x`. This means terms with `dy/dx` are `x^2 cos y dy/dx - 2e^(2y) cos x dy/dx`. So, `(x^2 cos y - 2e^(2y) cos x) dy/dx`. And terms without `dy/dx` are `6x + 2 - 2x sin y - e^(2y) sin x`. Therefore, `dy/dx = (6x + 2 - 2x sin y - e^(2y) sin x) / (x^2 cos y - 2e^(2y) cos x)`. If we check option 'a' in the provided image (page 7, 52): `dy/dx = (6x - 2x sin y - sin x e^(2y) + 2) / (x^2 cos y + 2 cos x e^(2y))` My numerator is `6x + 2 - 2x sin y - e^(2y) sin x`. This matches. My denominator is `x^2 cos y - 2e^(2y) cos x`. The option's denominator is `x^2 cos y + 2 cos x e^(2y)`. There is a sign difference. However, often in multiple choice, sometimes a negative sign on both numerator and denominator can make it match. Let's multiply my `dy/dx` by `-1/-1`: `dy/dx = -(6x + 2 - 2x sin y - e^(2y) sin x) / -(x^2 cos y - 2e^(2y) cos x)` `dy/dx = (-6x - 2 + 2x sin y + e^(2y) sin x) / (-x^2 cos y + 2e^(2y) cos x)` This doesn't seem to match the options either. Let's assume the provided option `a` is the intended answer and there might be a very subtle error in transcription or a specific context not visible. If the question implies `+e^(2y)cosx` instead of `-e^(2y)cosx`, then the derivative would be: `d/dx(e^(2y) cos x) = (2e^(2y) dy/dx) cos x + e^(2y) (-sin x)` So, `2x sin y + x^2 cos y dy/dx + 2e^(2y) cos x dy/dx - e^(2y) sin x = 6x + 2` ` (x^2 cos y + 2e^(2y) cos x) dy/dx = 6x + 2 - 2x sin y + e^(2y) sin x` ` dy/dx = (6x + 2 - 2x sin y + e^(2y) sin x) / (x^2 cos y + 2e^(2y) cos x)` This is close to option `a` if `e^(2y) sin x` term in numerator also has a negative. Given the image highlights option `a`, I will state it and acknowledge the discrepancy. Correct Option (based on selection in image): Option a. `dy/dx = (6x - 2x sin y - sin x e^(2y) + 2) / (x^2 cos y + 2 cos x e^(2y))` *Note: There seems to be a discrepancy in the sign of the `e^(2y) cos x` term's derivative in the denominator compared to standard calculation.*

1.6. Derivative of log(x^3 - x) + 7^(3/2)x

Question: Differentiate `y = log(x^3 - x) + 7^(3/2)x` (Appears as Question 7, 10, 2, 9 in different pages).

Solution: Assume `log` is natural logarithm `ln`.

`d/dx [ln(x^3 - x) + 7^(3/2)x]`

1. `d/dx(ln(x^3 - x))`: (Chain rule) `= 1/(x^3 - x) * d/dx(x^3 - x)` `= (3x^2 - 1) / (x^3 - x)` `= (3x^2 - 1) / (x(x^2 - 1))`

2. `d/dx(7^(3/2)x)`: This is of the form `a^(bx)`. The derivative of `a^u` is `a^u * ln a * u'`. Here `a=7`, `u=(3/2)x`. So `u' = 3/2`. `= 7^(3/2)x * ln 7 * (3/2)`

Combine the terms:

`dy/dx = (3x^2 - 1) / (x(x^2 - 1)) + (3/2) * 7^(3/2)x * ln 7`

The options typically write `ln 7` as `log_e 7` or similar. The options often use `(2/3)log 7` which is incorrect because it should be `(3/2)ln 7`. Let's assume there's a typo in `(2/3)log7` in the options and it should be `(3/2)ln7`. The options show `7^(3/2)x (2/3log7)`. This usually implies `7^(3/2)x * ln(7^(2/3))`. This is strange. Let's re-evaluate `d/dx(a^x) = a^x ln a`. If the term is `7^(3/2 * x)`, then `d/dx(7^(3x/2)) = 7^(3x/2) * ln 7 * (3/2)`. If the term is `7^(3/2) * x`, then `d/dx(7^(3/2) * x) = 7^(3/2)`. This is a constant multiplied by `x`. Given the options, it seems to be `7^(3x/2)`. The structure `7^(3/2)x` is ambiguous. If it means `(7^(3/2))*x`, the derivative is just `7^(3/2)`. If it means `7^((3/2)x)`, then the derivative is `7^((3/2)x) * ln 7 * (3/2)`. Looking at option `d` which is selected: `d. (3x^2 - 1) / (x(x^2 - 1)) + 7^(3/2)x (2/3 log7)` The `(2/3 log7)` part is still problematic. `ln 7` is a specific number. `(2/3)log 7` implies `(2/3) * ln 7`. Let's assume the `(2/3 log7)` is a way to represent the constant `(3/2) ln 7` for some reason, or there's a typo in the question's exponential term or derivative of exponential form in options. If the exponential term was `7^(2x/3)`, then its derivative would be `7^(2x/3) * ln 7 * (2/3)`. It's `7^(3/2)x` and option `d` has `7^(3/2)x (2/3log7)`. This is very likely a typo in the options or question formulation of the exponential term. However, the logarithmic part `(3x^2 - 1) / (x(x^2 - 1))` is consistent. Assuming the first term is correct and the second term is the one given in option `d` (despite the `log7` issue, which usually should be `ln 7`), we proceed. Correct Option (based on selection in image): Option d. `(3x^2 - 1) / (x(x^2 - 1)) + 7^(3/2)x (2/3 log7)` *Note: The derivative of `7^((3/2)x)` should be `(3/2) * 7^((3/2)x) * ln 7`. The `(2/3 log7)` term in the options is mathematically inconsistent with this, suggesting a typo in the question or options.*

1.7. Derivative of e^(sin 2x) / cos 2x

Question: If `y = e^(sin 2x) / cos 2x`, find `dy/dx` (Appears as Question 3, 4 in different pages).

Solution: This requires the quotient rule: `d/dx(u/v) = (u'v - uv') / v^2` Let `u = e^(sin 2x)` and `v = cos 2x`.

Find `u'` and `v'`:

• `u' = d/dx(e^(sin 2x))` = `e^(sin 2x) * d/dx(sin 2x)` = `e^(sin 2x) * (cos 2x * d/dx(2x))` = `2 cos 2x * e^(sin 2x)`
• `v' = d/dx(cos 2x)` = `-sin 2x * d/dx(2x)` = `-2 sin 2x`

Apply the quotient rule:

`dy/dx = ( (2 cos 2x * e^(sin 2x)) * cos 2x - e^(sin 2x) * (-2 sin 2x) ) / (cos 2x)^2`

`dy/dx = ( 2 cos^2 2x * e^(sin 2x) + 2 sin 2x * e^(sin 2x) ) / cos^2 2x`

Factor out `2e^(sin 2x)` from the numerator:

`dy/dx = ( 2e^(sin 2x) * (cos^2 2x + sin 2x) ) / cos^2 2x`

This does not simplify to `tan 2x` or `sec 2x` easily from `cos^2 2x + sin 2x`. Let's try to split the fraction instead:

`dy/dx = 2e^(sin 2x) * (cos^2 2x / cos^2 2x + sin 2x / cos^2 2x)` `dy/dx = 2e^(sin 2x) * (1 + (sin 2x / cos 2x) * (1 / cos 2x))` `dy/dx = 2e^(sin 2x) * (1 + tan 2x * sec 2x)`

This matches option `c` (or a variation of it, e.g., `sec 2x tan 2x`).

Correct Option: Option c. `dy/dx = 2e^(sin 2x) * [1 + sec 2x tan 2x]`

1.8. Derivative of log_e (cos x)

Question: Find `dy/dx` if `y = log_e (cos x)` (Appears as Question 9, 6 in different pages).

Solution: `y = ln(cos x)` Using the chain rule: `d/dx(ln u) = (1/u) * u'` Here, `u = cos x`, so `u' = d/dx(cos x) = -sin x`.

`dy/dx = (1 / cos x) * (-sin x)` `dy/dx = -sin x / cos x` `dy/dx = -tan x`

Correct Option: Option c. `-tanx`

2. Integration Questions

2.1. Integrate Cos(ln x)/x dx

Question: Evaluate `∫ Cos(ln x)/x dx` (Appears as Question 5, 4 in different pages).

Solution: This is a substitution integral. Let `u = ln x`. Then `du = (1/x) dx`. Substitute `u` and `du` into the integral:

`∫ Cos(u) du`

Integrate `Cos(u)` with respect to `u`:

`= sin u + C`

Substitute back `u = ln x`:

`= sin(ln x) + C`

Correct Option: Option d. `sin(ln x) + C`

2.2. Integrate 3x^2 e^(x^3) dx

Question: Simplify `∫ 3x^2 e^(x^3) dx` (Appears as Question 4, 3 in different pages).

Solution: This is a substitution integral. Let `u = x^3`. Then `du = d/dx(x^3) dx = 3x^2 dx`. Substitute `u` and `du` into the integral:

`∫ e^u du`

Integrate `e^u` with respect to `u`:

`= e^u + C`

Substitute back `u = x^3`:

`= e^(x^3) + C`

Correct Option: Option b. `e^(x^3) + C`

2.3. Integrate (4x-5) / (x^2-x-2) dx

Question: Simplify `∫ (4x-5) / (x^2-x-2) dx` (Appears as Question 3, 1, 5 in different pages).

Solution: This requires partial fraction decomposition.

First, factor the denominator: `x^2 - x - 2 = (x-2)(x+1)`

Set up the partial fractions:

`(4x-5) / ((x-2)(x+1)) = A/(x-2) + B/(x+1)`

Multiply both sides by `(x-2)(x+1)`:

`4x - 5 = A(x+1) + B(x-2)`

To find A, set `x = 2`:

`4(2) - 5 = A(2+1) + B(2-2)` `8 - 5 = 3A + 0` `3 = 3A` `A = 1`

To find B, set `x = -1`:

`4(-1) - 5 = A(-1+1) + B(-1-2)` `-4 - 5 = 0 + B(-3)` `-9 = -3B` `B = 3`

So the integral becomes:

`∫ [1/(x-2) + 3/(x+1)] dx`

Integrate term by term:

`∫ (1/(x-2)) dx + ∫ (3/(x+1)) dx` `= ln|x-2| + 3 ln|x+1| + C`

Correct Option: Option b. `3 ln|x+1| + ln|x-2| + C`

2.4. Integrate (e^(2x) + e^(-2x)) / (e^(2x) - e^(-2x)) dx

Question: Evaluate `∫ (e^(2x) + e^(-2x)) / (e^(2x) - e^(-2x)) dx` (Appears as Question 7, 8 in different pages).

Solution: This is a substitution integral. Let `u = e^(2x) - e^(-2x)`. Then `du = d/dx(e^(2x) - e^(-2x)) dx` `du = (2e^(2x) - (-2e^(-2x))) dx` `du = (2e^(2x) + 2e^(-2x)) dx` `du = 2(e^(2x) + e^(-2x)) dx`

So, `(e^(2x) + e^(-2x)) dx = (1/2) du`.

Substitute `u` and `du` into the integral:

`∫ (1/u) * (1/2) du` `= (1/2) ∫ (1/u) du` `= (1/2) ln|u| + C`

Substitute back `u = e^(2x) - e^(-2x)`:

`= (1/2) ln|e^(2x) - e^(-2x)| + C`

Correct Option: Option c. `(1/2) ln|e^(2x) - e^(-2x)| + C`

2.5. Integrate 1 / (x^2 + 4x + 7) dx

Question: Evaluate `∫ 1 / (x^2 + 4x + 7) dx` (Appears as Question 8).

Solution: This involves completing the square in the denominator and using the `arctan` integral form: `∫ 1/(a^2 + u^2) du = (1/a) arctan(u/a) + C`.

Complete the square for `x^2 + 4x + 7`:

`x^2 + 4x + (4/2)^2 - (4/2)^2 + 7` `= x^2 + 4x + 4 - 4 + 7` `= (x+2)^2 + 3`

Now, the integral becomes:

`∫ 1 / ((x+2)^2 + 3) dx`

Let `u = x+2`, then `du = dx`. Let `a^2 = 3`, so `a = sqrt(3)`.

`∫ 1 / (u^2 + a^2) du` `= (1/a) arctan(u/a) + C`

Substitute back `u = x+2` and `a = sqrt(3)`:

`= (1/sqrt(3)) arctan((x+2)/sqrt(3)) + C`

Correct Option: Option b. `(1/sqrt(3)) tan^-1((x+2)/sqrt(3)) + C`

2.6. Integrate Cos^3 x dx

Question: Evaluate `∫ Cos^3 x dx` (Appears as Question 5).

Solution: Use the identity `Cos^2 x = 1 - Sin^2 x`.

`∫ Cos^3 x dx = ∫ Cos^2 x * Cos x dx` `= ∫ (1 - Sin^2 x) * Cos x dx`

Let `u = Sin x`. Then `du = Cos x dx`.

`= ∫ (1 - u^2) du`

Integrate term by term:

`= u - (u^3)/3 + C`

Substitute back `u = Sin x`:

`= Sin x - (Sin^3 x)/3 + C`

Correct Option: Option a. `Sin x - (1/3) Sin^3 x + C`

3. Concepts and Definitions

3.1. Definition of an Indefinite Integral

Question: If `y = ∫f(x)dx + C`, then `∫f(x)dx` is called the _____ (Appears as Question 9, 8 in different pages).

Answer: The expression `∫f(x)dx` represents the indefinite integral of `f(x)`. It denotes the family of all antiderivatives of `f(x)`, where `C` is the constant of integration.

Correct Option: Option c. indefinite integral

3.2. Properties of Integrals

Question: Which of the following equation is not correct? (Appears as Question 1).

The options were not fully visible in the image, but one typical incorrect statement regarding integrals is about linearity or bounds. For example:

• `∫f(x)dx = 0` (Incorrect, unless `f(x)=0` or specific definite integral from a to a)
• `∫f(x)dx = ∫f(x)dx` (Correct, tautology)
• `∫f(x)dx ≤ ∫g(x)dx` if `f(x) ≤ g(x)` (Correct property for definite integrals over an interval)
• `∫f(x)dx ≤ ∫f(x)dx` (Correct)

Based on the partially visible options (e.g., `∫f(x)dx = 0`), the incorrect statement would be one that violates fundamental properties of integrals.

Correct Option (Likely): Option a. `∫f(x)dx = 0` (This is only true for specific cases, not a general property for indefinite integrals or for all definite integrals)