Category Archives: Calculus

Stewart – Calculus – 3.5 – Implicit Differentiation and Derivatives of Inverse Trigonometric Functions




Find y'' by implicit differentiation:


y''=-9(\frac{y\cdot 1\text{--}x\cdot y'}{y^{2}})

y''=-9(\frac{y\cdot 1\text{--}x(-9x/y)}{y^{2}}) (y' is replaced with -9x/y)

y''=-9(\frac{y^{2}+9x^{2}}{y^{3}}) (after multiplying by y to eliminate denominator from -9x)

y''=-9(\frac{9}{y^{3}}) (notice that y^{2}+9x^{2} equals the original equation)

So, y''=\frac{-81}{y^{3}}.

Derivatives of Inverse Trigonometric Functions


Derivatives of Inverse Trigonometric Functions

Stewart – Calculus – 3.3 – Derivatives of Trigonometric Functions

\frac{d}{\mathit{dx}}(\sin x)=? \frac{d}{\mathit{dx}}(\cos x)=? \frac{d}{\mathit{dx}}(\tan x)=? \frac{d}{\mathit{dx}}(\mathrm{csc} x)=? \frac{d}{\mathit{dx}}(\mathrm{sec} x)=? \frac{d}{\mathit{dx}}(\mathrm{cot} x)=?
To learn the derivatives of the 6 trig functions  you actually only have to learn 3 of them. The three to learn are sine, tangent, and secant.

F(X) = sin X
F'(X) = cos X
F(X) = tan X
F'(X) = sec2 X
F(X) = sec X
F'(X) = sec X tan X

The other three functions all start with "co": 
Cosine, Cotangent, and Cosecant. 
Think of the functions as having partners:


To find the derivative of the "co" functions, start with the derivatives of sine, tangent and secant, change each function in the derivative to it’s co-function partner and put a minus sign in front of the derivative.
So starting with one of the three we need to know:

If F(x) = secant X 
then F'(x) = secant X tangent X

 So now to get the derivative of the cosecant:
 If F(X) = cosecant X 
then F'(X) = -cosecant X cotangent X

So all 6 trig function derivatives look like this:

F(X) = sinX
F'(X) = cosX
F(X) = cosX
F'(X) = – sinX
F(X) = tanX
F'(X) = sec2 X
F(X) = cotX
F'(X) = – csc2 X
F(X) = secX
F'(X) = sec X tan X
F(X) = cscX
F'(X) = -csc X cot X

Thanks to Bruce Kirkpatrick for this.

Stewart – Calculus – 3.2 – Differentiation Rules




(a) If f(x)=xe^{x}, find f'(x).

(b) Find the nth derivative, f^{(n)}(x).

Differentiate the function f(t)=\sqrt{t}(a+\mathit{bt})

If f(x)=\sqrt{x} \hspace{1 mm} g(x), where g(4)=2 and g'(4)=3, find f'(4).



Let y=\frac{x^{2}+x\text{--}2}{x^{3}+6}

Find an equation of the tangent line to the curve y=\frac{e^{x}}{(1+x^{2})} at the point (1,\frac{1}{2}e).

 \frac{d}{\mathit{dx}}(c)=?   \frac{d}{\mathit{dx}}(x^{n})=?   \frac{d}{\mathit{dx}}(e^{x})=?   (\mathit{cf})'=?   (f+g)'=?   (f-g)'=?   (\mathit{fg})'=?   (\frac{f}{g})'=?
Using the Product Rule

If we first use the laws of exponents to rewrite f(t), then we can proceed directly without using the Product Rule.


Stewart – Calculus – 2.5 – Continuity

  1. Definition: A function f is continuous at a number a if

    This definition implicitly requires three things if f is continuous at a:

    1. f(a) is defined (that is, a is in the domain of f)

    2. \lim_{x \to a}f(x) exists

    3. \lim_{x \to a}f(x)=f(a)

  2. If f and g are continuous functions with f(3)=5 and
    \lim_{x \to 3}[2f(x) - g(x)]=4, find g(3).

    Use the definition of continuity and the properties of limits to show that the funtion is continous at the given number a. f(x)=(x+2x^{3})^{4}, a=-1

  3. The Intermediate Value Theorem
    Suppose that f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a) \neq f(b). Then there exists a number c in (a,b) such that f(c)=N.

    ~ ~ ~

    Use the Intermediate Value Theorem to show that there is
    a root of the given equation in the specified interval.

    x^4+x-3=0, (1,2)

Calculus Test 1

  1. Specify the domain of the function y = |x – 1|.

  2. Which of the following is a true statement about the graph of the equation y = x^4 +1?
  3. 1. It is symmetric about the x-axis.
    2. It is symmetric about the y-axis.
    3. It has two x-intercepts.
    4. It has no x-intercepts.

    a. Only statements 2 and 3 are true.
    b. Only statements 2 and 4 are true.
    c. Only statements 1 and 3 are true.
    d. Only statement 2 is true.

  4. Which of the following statements are true of the graph of y = \frac{2x-1}{x+1} ?

    1. It has no x-intercept.
    2. It has a slant asymptote y=2x.
    3. It has a vertical asymptote at x=-1.
    4. It has a horizontal asymptote at y=2.

    a. Statements 1, 2, and 3 are true.
    b. Statements 3 and 4 are true.
    c. Statements 2, 3, and 4 are true.
    d. All four statements are true.

  5. Let f(x) = 2 \cos x. The domain of f^{-1}(x) is

    1. [-1,1] 2. (2,  \infty) 3. (-\infty,  \infty) 4. (-2,2) 5. [-2,2]

  6. Simplify as far as possible. \ln e +a^{\log_{a}5}- \log100+10^0 - \log_{3} \frac{1}{3}
  7. Solve for x.

    a. \log_3 x + \log_3(2x+5)=1

    b. \frac{1}{\sqrt{2}}=4^{x}

  8. f(x)=(x-3)(x+1)^2(x-1)^4
  9. Identify the parts of the following composite function. f(g(h(j(x))))=\frac{1}{\sqrt{\log(x-1)}}
  10. Consider the picture below. Find the equation of the line L.

  11. The following table represents a function of the form f(x)=ab^{x}. Find the equation of the function.

     \begin{tabular}{| l | c | r |}  \hline  x & f(x)\\ \hline    0 & 6\\ \hline    1 & 18\\ \hline    2 & 54\\ \hline    3 & 162\\ \hline    4 & 486\\  \hline  \end{tabular}
  12. Given a^{m}=2, a^{n}=3, b^{m}=4, and b^{n}=5,
    use the properties of exponentials to determine (a^{3n}b^{m+n})^{\frac{1}{3}}.

  13. Evaluate:
    1. \sin \frac{\pi}{3}
    2. \tan \frac{3\pi}{4}
    3. \cos \frac{5\pi}{6}
  14. Find all solutions to the equaion, \tan^{2}x=\tan x, such that x \in [0,2\pi].
    Express the answers in radians.

  15. Find f^{-1}(x) if f(x) =  \sqrt{e^{x}+2}.
all real numbers
1. Set x equal to 0 to find the y-intercept.
2. Set the values in parentheses equal to 0 to get the x-intercepts 3, —1, and 1.
4. Note that the function will not cross the x axis with an even power, but will cross with an odd power.