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Calculus I Derivatives and Curve Sketching Cheat Sheet with Practice Problems

Calculus I Derivatives and Curve Sketching Cheat Sheet

This sheet covers the main ideas from the test review and adds practice problems with step-by-step solutions.

1) Derivative Basics and Rules

What to know

The derivative measures instantaneous rate of change.

It is the slope of the tangent line.

Common rules:

Power rule: \(\frac{d}{dx}(x^n)=nx^{n-1}\)

Constant rule: derivative of a constant is 0

Constant multiple rule: pull constants out

Sum/difference rule: differentiate term by term

Product rule: \((fg)'=f'g+fg'\)

Quotient rule: \(\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}\)

Chain rule: \(\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\)

Practice Problem 1

Differentiate \(f(x)=3x^4-5x^2+7\).

Step-by-step

1.

Differentiate each term separately.

2.

Use the power rule on the powers of \(x\).

3.

The constant \(7\) becomes 0.

Solution

\[

f'(x)=12x^3-10x

\]

Practice Problem 2

Differentiate \(g(x)=(x^2+1)(x^3-4)\).

Step-by-step

1.

Identify the two functions: \(u=x^2+1\), \(v=x^3-4\).

2.

Compute derivatives: \(u'=2x\), \(v'=3x^2\).

3.

Apply the product rule: \(u'v+uv'\).

4.

Simplify.

Solution

\[

g'(x)=2x(x^3-4)+(x^2+1)(3x^2)

\]

\[

g'(x)=2x^4-8x+3x^4+3x^2=5x^4+3x^2-8x

\]

Practice Problem 3

Differentiate \(h(x)=\frac{x^2+1}{x-3}\).

Step-by-step

1.

Let \(u=x^2+1\), \(v=x-3\).

2.

Compute \(u'=2x\), \(v'=1\).

3.

Use the quotient rule.

4.

Simplify carefully.

Solution

\[

h'(x)=\frac{(2x)(x-3)-(x^2+1)(1)}{(x-3)^2}

\]

\[

h'(x)=\frac{2x^2-6x-x^2-1}{(x-3)^2}=\frac{x^2-6x-1}{(x-3)^2}

\]

Practice Problem 4

Differentiate \(p(x)=(4x^2-1)^5\).

Step-by-step

1.

Recognize the outside function is a power.

2.

Use the chain rule.

3.

Differentiate the inside function.

Solution

\[

p'(x)=5(4x^2-1)^4(8x)=40x(4x^2-1)^4

\]

2) Trigonometric Derivatives and Unit Circle Values

What to know

\((\sin x)'=\cos x\)

\((\cos x)'=-\sin x\)

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

You may need unit circle values such as:

\(\sin\left(\frac{\pi}{6}\right)=\frac12\)

\(\cos\left(\frac{\pi}{3}\right)=\frac12\)

\(\tan\left(\frac{\pi}{4}\right)=1\)

Practice Problem 5

Differentiate \(f(x)=\sin x+3\cos x\).

Step-by-step

1.

Differentiate \(\sin x\) to get \(\cos x\).

2.

Differentiate \(3\cos x\) to get \(-3\sin x\).

3.

Combine the results.

Solution

\[

f'(x)=\cos x-3\sin x

\]

Practice Problem 6

Evaluate \(\tan\left(\frac{\pi}{4}\right)\).

Step-by-step

1.

Use the unit circle or special triangle values.

2.

Recall \(\sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\).

3.

Compute \(\tan x=\frac{\sin x}{\cos x}\).

Solution

\[

\tan\left(\frac{\pi}{4}\right)=1

\]

3) Inverse Functions and Inverse Trig Derivatives

What to know

To find an inverse function:

1.

Replace \(f(x)\) with \(y\)

2.

Swap \(x\) and \(y\)

3.

Solve for \(y\)

4.

Write \(f^{-1}(x)\)

Common inverse trig derivative formulas:

\(\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}\)

\(\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}\)

Practice Problem 7

Find the inverse of \(f(x)=2x-5\).

Step-by-step

1.

Write \(y=2x-5\).

2.

Swap \(x\) and \(y\): \(x=2y-5\).

3.

Solve for \(y\).

Solution

\[

x=2y-5

\]

\[

x+5=2y

\]

\[

y=\frac{x+5}{2}

\]

So,

\[

f^{-1}(x)=\frac{x+5}{2}

\]

Practice Problem 8

Differentiate \(f(x)=\arctan x\).

Step-by-step

1.

Recall the inverse trig derivative formula.

2.

Substitute the function directly.

Solution

\[

f'(x)=\frac{1}{1+x^2}

\]

4) Implicit Differentiation

What to know

Use implicit differentiation when \(y\) is mixed with \(x\).

Differentiate both sides with respect to \(x\).

Every time you differentiate a term containing \(y\), multiply by \(\frac{dy}{dx}\).

For second derivatives, differentiate again after solving for \(\frac{dy}{dx}\) if needed.

Practice Problem 9

Find \(\frac{dy}{dx}\) for \(x^2+y^2=25\).

Step-by-step

1.

Differentiate both sides with respect to \(x\).

2.

Remember \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\).

3.

Solve for \(\frac{dy}{dx}\).

Solution

\[

2x+2y\frac{dy}{dx}=0

\]

\[

2y\frac{dy}{dx}=-2x

\]

\[

\frac{dy}{dx}=-\frac{x}{y}

\]

Practice Problem 10

Find \(\frac{dy}{dx}\) for \(x^2+xy+y^2=7\).

Step-by-step

1.

Differentiate each term.

2.

Use the product rule on \(xy\).

3.

Collect the \(\frac{dy}{dx}\) terms.

4.

Solve.

Solution

\[

2x+(x\frac{dy}{dx}+y)+2y\frac{dy}{dx}=0

\]

\[

(x+2y)\frac{dy}{dx}=-(2x+y)

\]

\[

\frac{dy}{dx}=-\frac{2x+y}{x+2y}

\]

5) Logarithmic and Exponential Differentiation

What to know

\((e^x)'=e^x\)

\((a^x)'=a^x\ln a\)

\((\ln x)'=\frac{1}{x}\)

\((\log_a x)'=\frac{1}{x\ln a}\)

Logarithmic differentiation is useful when variables appear in both base and exponent.

Practice Problem 11

Differentiate \(f(x)=e^{3x}\).

Step-by-step

1.

Use the chain rule.

2.

Keep the exponential function.

3.

Multiply by the derivative of the inside.

Solution

\[

f'(x)=3e^{3x}

\]

Practice Problem 12

Differentiate \(f(x)=\ln(x^2+1)\).

Step-by-step

1.

Use the chain rule.

2.

Derivative of \(\ln(u)\) is \(\frac{u'}{u}\).

3.

Here \(u=x^2+1\), so \(u'=2x\).

Solution

\[

f'(x)=\frac{2x}{x^2+1}

\]

Practice Problem 13

Differentiate \(y=x^x\).

Step-by-step

1.

Take natural logs of both sides: \(\ln y=x\ln x\).

2.

Differentiate both sides implicitly.

3.

Solve for \(\frac{dy}{dx}\).

4.

Multiply by \(y\), then replace \(y\) with \(x^x\).

Solution

\[

\frac{1}{y}\frac{dy}{dx}=\ln x+1

\]

\[

\frac{dy}{dx}=y(\ln x+1)

\]

\[

\frac{dy}{dx}=x^x(\ln x+1)

\]

6) Related Rates

What to know

Related rates problems connect two or more changing quantities.

Main steps:

1.

Draw a picture.

2.

Write an equation relating the variables.

3.

Differentiate with respect to time \(t\).

4.

Plug in known values.

5.

Solve for the unknown rate.

Practice Problem 14

A circle’s radius is increasing at \(\frac{dr}{dt}=2\) cm/s. Find \(\frac{dA}{dt}\) when \(r=5\).

Step-by-step

1.

Use the area formula \(A=\pi r^2\).

2.

Differentiate with respect to time.

3.

Substitute the given values.

Solution

\[

\frac{dA}{dt}=2\pi r\frac{dr}{dt}

\]

\[

\frac{dA}{dt}=2\pi(5)(2)=20\pi

\]

So,

\[

\frac{dA}{dt}=20\pi\text{ cm}^2/\text{s}

\]

Practice Problem 15

A balloon’s volume is increasing. If \(V=\frac{4}{3}\pi r^3\) and \(\frac{dr}{dt}=3\), find \(\frac{dV}{dt}\) when \(r=4\).

Step-by-step

1.

Differentiate the volume formula with respect to time.

2.

Use the chain rule.

3.

Plug in \(r=4\) and \(\frac{dr}{dt}=3\).

Solution

\[

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

\]

\[

\frac{dV}{dt}=4\pi(4^2)(3)=192\pi

\]

So,

\[

\frac{dV}{dt}=192\pi\text{ cm}^3/\text{s}

\]

7) Optimization, Critical Values, and Extreme Values

What to know

Critical numbers occur where \(f'(x)=0\) or \(f'(x)\) does not exist.

To find absolute extrema on a closed interval, use the closed interval method:

1.

Find critical numbers in the interval.

2.

Check endpoints.

3.

Evaluate \(f(x)\) at each candidate.

4.

Compare values.

Practice Problem 16

Find the critical numbers of \(f(x)=x^3-3x\).

Step-by-step

1.

Differentiate the function.

2.

Set \(f'(x)=0\).

3.

Solve.

Solution

\[

f'(x)=3x^2-3=3(x^2-1)=3(x-1)(x+1)

\]

So the critical numbers are

\[

x=-1,\;1

\]

Practice Problem 17

Find the absolute max and min of \(f(x)=x^2-4x+1\) on \([0,4]\).

Step-by-step

1.

Find \(f'(x)\).

2.

Find critical numbers in \([0,4]\).

3.

Evaluate the function at critical numbers and endpoints.

4.

Compare values.

Solution

\[

f'(x)=2x-4

\]

Set equal to 0:

\[

2x-4=0 \Rightarrow x=2

\]

Now evaluate:

\[

f(0)=1,\quad f(2)=4-8+1=-3,\quad f(4)=16-16+1=1

\]

So the absolute minimum is \(-3\) at \(x=2\), and the absolute maximum is \(1\) at \(x=0\) and \(x=4\).

8) Rolle’s Theorem and the Mean Value Theorem

What to know

Rolle’s Theorem: If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a)=f(b)\), then there is some \(c\in(a,b)\) with \(f'(c)=0\).

Mean Value Theorem: If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is some \(c\in(a,b)\) with

\[

f'(c)=\frac{f(b)-f(a)}{b-a}

\]

Practice Problem 18

For \(f(x)=x^2\) on \([0,2]\), find the MVT value of \(c\).

Step-by-step

1.

Compute the average rate of change.

2.

Set \(f'(c)\) equal to that value.

3.

Solve for \(c\).

Solution

\[

\frac{f(2)-f(0)}{2-0}=\frac{4-0}{2}=2

\]

Since \(f'(x)=2x\), set

\[

2c=2

\]

\[

c=1

\]

Practice Problem 19

Use Rolle’s Theorem on \(f(x)=x^2-4x+3\) on \([1,3]\).

Step-by-step

1.

Check that \(f(1)=f(3)\).

2.

Find \(f'(x)\).

3.

Set \(f'(c)=0\) and solve.

Solution

\[

f(1)=1-4+3=0,\quad f(3)=9-12+3=0

\]

So Rolle’s Theorem applies.

\[

f'(x)=2x-4

\]

\[

2c-4=0 \Rightarrow c=2

\]

9) Curve Sketching: Increasing, Decreasing, Concavity, and Inflection Points

What to know

If \(f'(x)>0\), the function is increasing.

If \(f'(x)<0\), the function is decreasing.

If \(f''(x)>0\), the function is concave up.

If \(f''(x)<0\), the function is concave down.

Inflection points happen where concavity changes.

Practice Problem 20

Find where \(f(x)=x^3-3x^2\) is increasing and decreasing.

Step-by-step

1.

Find \(f'(x)\).

2.

Find critical numbers by setting \(f'(x)=0\).

3.

Test intervals using sign analysis.

Solution

\[

f'(x)=3x^2-6x=3x(x-2)

\]

Critical numbers: \(x=0,2\)

Test intervals:

On \((-

∞,0)\), \(f'(x)>0\)

On \((0,2)\), \(f'(x)<0\)

On \((2,\infty)\), \(f'(x)>0\)

So the function is increasing on \((-

∞,0)\cup(2,\infty)\) and decreasing on \((0,2)\).

Practice Problem 21

Find the concavity and inflection points of \(f(x)=x^3-3x\).

Step-by-step

1.

Find \(f''(x)\).

2.

Set \(f''(x)=0\) to find possible inflection points.

3.

Test intervals around those points.

4.

Check whether concavity changes.

Solution

\[

f'(x)=3x^2-3

\]

\[

f''(x)=6x

\]

Set \(f''(x)=0\):

\[

6x=0 \Rightarrow x=0

\]

Check sign:

If \(x<0\), \(f''(x)<0\): concave down

If \(x>0\), \(f''(x)>0\): concave up

So there is an inflection point at \(x=0\). Since \(f(0)=0\), the inflection point is \((0,0)\).

Practice Problem 22

Use the second derivative test for \(f(x)=x^3-3x\) at a critical point.

Step-by-step

1.

Find critical numbers from \(f'(x)=0\).

2.

Compute \(f''(x)\).

3.

Evaluate \(f''\) at the critical number.

4.

Use the sign of \(f''\) to classify the point.

Solution

Critical numbers come from

\[

f'(x)=3x^2-3=0 \Rightarrow x=\pm1

\]

Now use \(f''(x)=6x\):

At \(x=1\), \(f''(1)=6>0\), so there is a local minimum.

At \(x=-1\), \(f''(-1)=-6<0\), so there is a local maximum.

10) Quick Problem-Solving Checklist

If the problem asks you to differentiate

Identify the rule: power, product, quotient, chain, trig, log, exponential, or implicit.

Simplify first if needed.

Differentiate carefully term by term.

Simplify your answer.

If the problem asks for extrema or behavior

Find \(f'(x)\) and/or \(f''(x)\).

Solve for critical numbers or inflection candidates.

Use a sign chart or test points.

State intervals clearly.

If the problem is a word problem

Draw a picture if possible.

Write one equation relating the variables.

Differentiate with respect to time if rates are involved.

Substitute known values at the end.

11) Fast Reference Formulas

\[

\frac{d}{dx}(x^n)=nx^{n-1}

\]

\[

\frac{d}{dx}(\sin x)=\cos x

\]

\[

\frac{d}{dx}(\cos x)=-\sin x

\]

\[

\frac{d}{dx}(\tan x)=\sec^2 x

\]

\[

\frac{d}{dx}(e^x)=e^x

\]

\[

\frac{d}{dx}(\ln x)=\frac{1}{x}

\]

\[

(fg)'=f'g+fg'

\]

\[

\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}

\]

\[

\frac{d}{dx}f(g(x))=f'(g(x))g'(x)

\]

\[

\text{MVT: } f'(c)=\frac{f(b)-f(a)}{b-a}

\]

\[

\text{Rolle's Theorem: } f'(c)=0

\]

12) What to Memorize Before the Test

Derivative rules

Trig derivative formulas

Inverse trig derivative formulas

Implicit differentiation steps

Log and exponential derivative rules

Related rates setup process

Critical numbers and absolute extrema steps

Rolle’s Theorem and MVT conditions

Increasing/decreasing and concavity tests

First and second derivative tests