Session 01 - Calculus 1 Review

Solution for Question 1

Here is the step-by-step breakdown for solving each part of the calculus problem based on the function:

(a) Find $f'(x)$

Concept: The first derivative represents the instantaneous rate of change or the slope of the function at any given point. To find it, we use the Power Rule ($\frac{d}{dx}[x^n] = n x^{n-1}$) and the Constant Rule ($\frac{d}{dx}[c] = 0$).

1. Differentiate each term of the function individually:

• The derivative of $x^2$ is $2x$.

• The derivative of $-4x$ is $-4$.

• The derivative of the constant $4$ is $0$.

2. Combine the terms to get your derivative function:

(b) Find an equation of the line tangent to the graph of $f(x)$ at $x = -1$

Concept: To write the equation of a straight line, you need a point $(x_1, y_1)$ and a slope ($m$). Once you have both, you plug them into the point-slope formula: $y - y_1 = m(x - x_1)$.

1. Find the y-coordinate (the point): Evaluate the original function $f(x)$ at $x = -1$.

This gives you the coordinate point $(-1, 9)$.

2. Find the slope ($m$): Evaluate the derivative function $f'(x)$ at $x = -1$.

3. Construct the equation: Substitute the point $(-1, 9)$ and slope $m = -6$ into the point-slope form.

(c) For what value(s) of $x$ does $f(x)$ have a slope of 0?

Concept: Because the derivative $f'(x)$ gives you the slope at any $x$, you can find where the slope is exactly $0$ by setting the derivative equal to zero and solving for $x$.

1. Set $f'(x) = 0$:

2. Isolate $x$:

(d) For what value(s) of $x$ does $f(x)$ have a negative instantaneous rate of change?

Concept: "Instantaneous rate of change" is simply another term for the derivative. For this rate to be negative, the derivative must be less than zero ($f'(x) < 0$).

1. Set up the inequality using your derivative:

2. Solve for $x$:

In interval notation, the instantaneous rate of change is negative on the interval $(-\infty, 2)$.

Here are the step-by-step solutions for finding the first derivative of each function.

(a) $f(x) = \frac{1}{x^2} - 5\sqrt{x} - 3 \ln x + \pi^4$

Step 1: Rewrite the function using exponents Before taking the derivative, convert fractions and radicals into power forms so you can easily use the Power Rule ($\frac{d}{dx}[x^n] = n x^{n-1}$).

Step 2: Differentiate term by term

• First term ($x^{-2}$): Bring the exponent down and subtract 1.

• Second term ($-5x^{1/2}$): Bring the exponent down and multiply.

• Third term ($-3\ln x$): The derivative of $\ln x$ is $\frac{1}{x}$.

• Fourth term ($\pi^4$): Don't let this trap you! Even though it looks fancy, $\pi^4$ is just a constant number, and the derivative of any constant is $0$.

(b) $y = \frac{x^3 - 4x^2 + x}{x - 2}$

Step 1: Identify the rule Because we have a function divided by another function, we must use the Quotient Rule:

Let's define our components:

• Top ($u$): $x^3 - 4x^2 + x \implies u' = 3x^2 - 8x + 1$

• Bottom ($v$): $x - 2 \implies v' = 1$

Step 2: Apply the formula

Step 3: Expand and simplify the numerator

1. Expand the first part: $(3x^2 - 8x + 1)(x - 2) = 3x^3 - 6x^2 - 8x^2 + 16x + x - 2 = 3x^3 - 14x^2 + 17x - 2$

2. Subtract the second part: $(3x^3 - 14x^2 + 17x - 2) - (x^3 - 4x^2 + x)$

3. Combine like terms: $2x^3 - 10x^2 + 16x - 2$

Final Answer:

(c) $g(x) = (x + \sin x) \tan x$

Step 1: Identify the rule This is a product of two distinct functions, so we need the Product Rule:

Let's break down the pieces:

• First ($u$): $x + \sin x \implies u' = 1 + \cos x$

• Second ($v$): $\tan x \implies v' = \sec^2 x$

Step 2: Assemble the parts

Step 3: Optional simplification Distributing the $\tan x$ in the first block can make it look a bit cleaner since $\cos x \cdot \tan x = \sin x$:

Final Answer:

(d) $y = \cos(e^x) - e^{x^5}$

Step 1: Identify the rule Both terms involve an "inner" and "outer" function, meaning we must apply the Chain Rule:

Step 2: Handle term one ($\cos(e^x)$)

• Outer function is $\cos(\text{blob}) \implies -\sin(\text{blob})$

• Inner function is $e^x \implies e^x$

• Putting it together: $-\sin(e^x) \cdot e^x = -e^x \sin(e^x)$

Step 3: Handle term two ($-e^{x^5}$)

• Outer function is $e^{\text{blob}} \implies e^{\text{blob}}$

• Inner function is $x^5 \implies 5x^4$

• Putting it together: $-e^{x^5} \cdot 5x^4 = -5x^4 e^{x^5}$

Final Answer:

Before we dive in, remember the golden rule of indefinite integrals: never forget to add the constant of integration ($+ C$) at the very end!

(a) $\int \left(x + \frac{2}{x} - \frac{3}{x^3} + \frac{4}{\sqrt[4]{x}}\right) dx$

Step 1: Rewrite the terms using exponents Just like with derivatives, it is much easier to integrate fractions and roots if we convert them into standard power forms ($x^n$).

• $\frac{2}{x}$ stays as $2 \cdot \frac{1}{x}$ because it follows a special rule.

• $-\frac{3}{x^3}$ becomes $-3x^{-3}$

• $\frac{4}{\sqrt[4]{x}}$ becomes $4x^{-1/4}$

Now, rewrite the whole integral:

Step 2: Integrate term-by-term We will use the Power Rule for Integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$) for most of these, and the Natural Log Rule ($\int \frac{1}{x} dx = \ln|x|$) for the second term.

• First term ($x^1$): Add 1 to the exponent, then divide by the new exponent.

• Second term ($2 \cdot \frac{1}{x}$): The integral of $\frac{1}{x}$ is always $\ln|x|$.

• Third term ($-3x^{-3}$): Add 1 to the exponent ($-3 + 1 = -2$), then divide by $-2$.

• Fourth term ($4x^{-1/4}$): Add 1 to the exponent ($-\frac{1}{4} + 1 = \frac{3}{4}$), then divide by $\frac{3}{4}$ (which is the same as multiplying by $\frac{4}{3}$).

(b) $\int (2x + 1)\sqrt[5]{x^2 + x} \, dx$

Step 1: Identify the method ($u$-substitution) Notice that we have a complicated "inner" function ($x^2 + x$) sitting inside a fifth root, and its exact derivative ($2x + 1$) is multiplying the rest of the expression. This is a textbook cue to use $u$-substitution.

Let's define our pieces:

• Let $u = x^2 + x$

• Take the derivative to find $du$: $du = (2x + 1) \, dx$

Step 2: Substitute $u$ and $du$ into the integral Replace $(x^2 + x)$ with $u$, and replace $(2x + 1)dx$ with $du$.

Step 3: Rewrite and integrate Convert the root into a fractional exponent so you can apply the Power Rule:

Add 1 to the exponent ($\frac{1}{5} + 1 = \frac{6}{5}$), and divide by $\frac{6}{5}$ (multiply by $\frac{5}{6}$):

Step 4: Substitute back to the original variable ($x$) Replace $u$ back with your original expression $(x^2 + x)$ to finish the job.

Final Answer:

Unlike indefinite integrals, definite integrals give us a specific numerical value (representing the net area under the curve between two boundaries) and do not need a $+ C$ at the end. We will use the Fundamental Theorem of Calculus:

where $F(x)$ is the antiderivative.

(a) $\int_{1}^{3} \left(3x^2 - \frac{1}{4}x^3\right) dx$

Step 1: Find the antiderivative $F(x)$ Integrate each term individually using the Power Rule ($\int x^n dx = \frac{x^{n+1}}{n+1}$):

• First term ($3x^2$): $\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$

• Second term ($-\frac{1}{4}x^3$): $\int -\frac{1}{4}x^3 \, dx = -\frac{1}{4} \cdot \frac{x^4}{4} = -\frac{1}{16}x^4$

So, our antiderivative function is:

Step 2: Evaluate at the upper limit ($x = 3$) Plug $3$ into your antiderivative:

Step 3: Evaluate at the lower limit ($x = 1$) Plug $1$ into your antiderivative:

Step 4: Subtract the lower bound value from the upper bound value ($F(3) - F(1)$)

Simplifying the fraction gives exactly 21.

(b) $\int_{0}^{2} \frac{1}{5x+1} dx$

Step 1: Use $u$-substitution to find the antiderivative Because the denominator contains a linear function ($5x + 1$), we need to use a simple $u$-sub.

• Let $u = 5x + 1$

• Take the derivative: $du = 5 \, dx \implies dx = \frac{1}{5} \, du$

Step 2: Change the limits of integration to match $u$ When dealing with definite integrals, it's easiest to convert the $x$-boundaries into $u$-boundaries using your substitution formula ($u = 5x + 1$):

• Lower limit ($x = 0$): $u = 5(0) + 1 = 1$

• Upper limit ($x = 2$): $u = 5(2) + 1 = 11$

Step 3: Rewrite and evaluate the integral in terms of $u$ Substitute your components and the new boundaries into the integral:

The integral of $\frac{1}{u}$ is $\ln|u|$:

Step 4: Apply the boundaries

Since $\ln(1) = 0$, this simplifies beautifully:

Final Answer: