June, 2008

Instruction: Find the first derivative of the given function.

26. $y = \sec^3 \beta \, \tan^2 \beta$
    $\dfrac{dy}{d\beta} = \sec^3 \beta \, (2 \tan \beta \, \sec^2 \beta) + \tan^2 \beta \, (3 \sec^2 \beta)(\sec \beta \, \tan \beta)$
    $\dfrac{dy}{d\beta} = 2 \tan \beta \, \sec^5 \beta + 3 \tan^3 \beta \, \sec^3 \beta$
    $\dfrac{dy}{d\beta} = \tan \beta \, \sec^3 \beta \, (2 \sec^2 \beta + 3 \tan^2 \beta)$
    $\dfrac{dy}{d\beta} = \tan \beta \, \sec^3 \beta \, [2 \sec^2 \beta + 3 (\sec^2 \beta - 1)]$
    $\dfrac{dy}{d\beta} = \tan \beta \, \sec^3 \beta \, [2 \sec^2 \beta + 3 \sec^2 \beta - 3]$
    $\dfrac{dy}{d\beta} = \tan \beta \, \sec^3 \beta \, (5 \sec^2 \beta - 3)$        answer

Instruction: Find the first derivative of the given function.

21. $y = x^2 \, sin \frac{1}{2}x$
    $y' = x^2 \left[ \left( \cos \frac{1}{2}x \right) \left( \frac{1}{2} \right) \right] + \sin \frac{1}{2}x \, [ 2x ]$
    $y' = \frac{1}{2} x^2 \, \cos \frac{1}{2}x + 2x \, \sin \frac{1}{2}x$        answer

Instruction: Find the first derivative of the given function.

16. $u = \sin^2 3t$
    $\dfrac{du}{dt} = 2 \sin 3t \, (\cos 3t)(3)$
    $\dfrac{du}{dt} = 6 \sin 3t \, \cos 3t$        answer

The figure shown below are two inverted equilateral triangles inscribed in a circle of radius 50 cm. Find the area of the shaded region.

Find the area of the regular five pointed star inscribe in a circle of radius 50 cm.

Instruction: Find the first derivative of the given function.

11. $y = 6 \sec 3x$
    $y' = 6 \sec 3x \, \tan 3x \dfrac{d}{dx}(3x)$
    $y' = 6 \sec 3x \, \tan 3x \, (3)$
    $y' = 18 \sec 3x \, \tan 3x$        answer

Situation - For the system of forces shown in Fig. Eng. Mech. 01.

Instruction: Find the first derivative of the given function.

6. $y = \csc 7x$
   $y' = -\csc 7x \, \cot 7x \dfrac{d}{dx}(7x)$
   $y' = -7 \csc 7x \, \cot 7x$        answer

Instruction: Find the first derivative of the given function.

1. $y = \sin 3x$
   $y' = \cos 3x \dfrac{d}{dx}(3x)$
   $y' = 3 \cos 3x$        answer

Trigonometric identities and formulas are basic requirements for this section. If u is a function of x, then

1. $\dfrac{d}{dx}(\sin u) = \cos u \dfrac{du}{dx}$
2. $\dfrac{d}{dx}(\cos u) = - \sin u \dfrac{du}{dx}$
3. $\dfrac{d}{dx}(\tan u) = \sec^2 u \dfrac{du}{dx}$
4. $\dfrac{d}{dx}(\cot u) = - \csc^2 u \dfrac{du}{dx}$
5. $\dfrac{d}{dx}(\sec u) = \sec u \tan u \dfrac{du}{dx}$
6. $\dfrac{d}{dx}(\csc u) = -\csc u \cot u \dfrac{du}{dx}$

Click a link below for some examples of Differentiation of Trigonometric Functions.

Instruction: Find the slope of the curve at the given point.

35. $y^2 = \dfrac{2a^3 \, x}{x^2 + a^2}$ at (a, a).

    Solution:
    $y^2 = \dfrac{2a^3 \, x}{x^2 + a^2}$

    $2y \, y' = \dfrac{(x^2 + a^2) 2a^3 - 2a^3 \, x \, (2x)}{(x^2 + a^2)^2}$

    $2y \, y' = \dfrac{2a^3 \, x^2 + 2a^5 - 4a^3 \, x^2}{(x^2 + a^2)^2}$

    $y' = \dfrac{a^5 - a^3 \, x^2}{y \, (x^2 + a^2)^2}$

    At point (a, a):
    $y' = \dfrac{a^5 - a^3 \, (a^2)}{a \, (a^2 + a^2)^2}$

    $y' = 0$        answer

Instruction: Find the slope of the curve at the given point.

31. x (x² – y²) = 3 at (–1, 2).

    Solution:
    x (2x – 2y y') + (x² – y²) (1)= 0
    2x² – 2xy y' + (x² – y²) = 0
    2x² + (x² – y²) = 2xy y'
    $y' = \dfrac{3x^2 - y^2}{2xy}$

    At point (-1, 2)
    $y' = \dfrac{3(-1)^2 - 2^2}{2(-1)(2)}$
    $y' = \frac{1}{4}$        answer

Instruction: Find the slope of the curve at the given point.

27. x² + y² – 12x + 4y – 5 = 0 at (0, 1).

    Solution:
    x² + y² – 12x + 4y – 5 = 0
    2x + 2y y' - 12 + 4y' = 0
    x + y y' - 6 + 2y' = 0
    (y + 2) y' = 6 - x
    $y' = \dfrac{6 - x}{y + 2}$

    At point (0, 1)
    $y' = \dfrac{6 - 0}{1 + 2}$
    $y' = 2$        answer

$$ \displaystyle \int \frac {2x \, dx}{\sqrt {4x^2+9}} $$

Use the Power Formula to integrate
$\displaystyle \int \sin x \cos x \, dx$ in two ways.

INSTRUCTION: Find the derivative of y with respect to x by using the given relation in its implicit form.

24. If x³ = at², find $\dfrac{d^2x}{dt^2}$.

    Solution 24
    x³ = at²
    $3x^2 \dfrac{dx}{dt} = 2at$

    $\dfrac{dx}{dt} = \dfrac{2at}{3x^2}$

    $\dfrac{d^2x}{dt^2} = \dfrac{3x^2 (2a) - 2at \left( 6x \dfrac{dx}{dt} \right)}{(3x^2)^2}$

    $\dfrac{d^2x}{dt^2} = \dfrac{6ax^2 - 12atx \dfrac{dx}{dt}}{9x^4}$

    $\dfrac{d^2x}{dt^2} = \dfrac{6ax^2 - 12atx \left( \dfrac{2at}{3x^2} \right)}{9x^4}$

    $\dfrac{d^2x}{dt^2} = \dfrac{6ax^2 - \dfrac{8a^2 t^2}{x}}{9x^4}$

    $\dfrac{d^2x}{dt^2} = \dfrac{\dfrac{6ax^3 - 8a^2 t^2}{x}}{9x^4}$

    $\dfrac{d^2x}{dt^2} = \dfrac{2a(3x^3 - 4a t^2)}{9x^3 \cdot x^2}$

    $\dfrac{d^2x}{dt^2} = \dfrac{2a(3a t^2 - 4a t^2)}{9at^2 x^2}$

    $\dfrac{d^2x}{dt^2} = \dfrac{2a(-a t^2)}{9at^2 x^2}$

    $\dfrac{d^2x}{dt^2} = -\dfrac{2a}{9x^2}$        answer

INSTRUCTION: Find the derivative of y with respect to x by using the given relation in its implicit form.

21. If x² + y² = a², find y''.

    Solution 21
    x² + y² = a²
    2x + 2y y' = 0
    x + y y' = 0
    $y' = -\dfrac{x}{y}$

    $y'' = -\dfrac{y(1) - xy'}{y^2}$

    $y'' = -\dfrac{y - x\left( -\dfrac{x}{y} \right)}{y^2}$

    $y'' = -\dfrac{y + \dfrac{x^2}{y}}{y^2}$

    $y'' = -\dfrac{\dfrac{y^2 + x^2}{y}}{y^2}$

    $y'' = -\dfrac{x^2 + y^2}{y^3}$

    $y'' = -\dfrac{a^2}{y^3}$        answer

INSTRUCTION: Find the derivative of y with respect to x by using the given relation in its implicit form.

16. If y = (x³ + 1)², find $\dfrac{dx}{dy}$.

    Solution 16
    y = (x³ + 1)²
    dy = 2(x³ + 1)(3x² dx)
    1 = 3x²(x³ + 1) $\dfrac{dx}{dy}$
    $\dfrac{dx}{dy} = \dfrac{1}{6x^2 (x^3 + 1)}$        answer

INSTRUCTION: Find the derivative of y with respect to x by using the given relation in its implicit form.

11. (x² – y²)² = 4ay³
    2(x² - y²)(2x - 2y y') = 12ay² y'
    4(x² - y²)(x - y y') = 12ay² y'
    (x² - y²)(x - y y') = 3ay² y'
    x (x² - y²) - y (x² - y²) y' = 3ay² y'
    x (x² - y²) = 3ay² y' + y (x² - y²) y'
    x (x² - y²) = y (3ay + x² - y²) y'
    $y' = \dfrac{x (x^2 - y^2)}{y (3ay + x^2 - y^2)}$        answer

INSTRUCTION: Find the derivative of y with respect to x by using the given relation in its implicit form.

6. x² – 2xy + y² – 6x + 2y = 0
   2x - 2(xy' + y) + 2yy' - 6 + 2y' = 0
   x - xy' - y + yy' - 3 + y' = 0
   (x - y - 3) - (x - y - 1)y' = 0
   (x - y - 1)y' = x - y - 3
   $y' = \dfrac{x - y - 3}{x - y - 1}$        answer

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INSTRUCTION: Find the derivative of y with respect to x by using the given relation in its implicit form.

1. x² + y² = a²
   2x + 2y y' = 0
   x + y y' = 0
   $y' = -\dfrac{x}{y}$        answer

We have been dealing in the previous section functions that are explicitly defined, that is, the dependent variable is expressed as a function of independent variable as of the function y = f(x). There are instances that a function of x and y is in the form F(x, y) = 0, neither x nor y is expressed as a function of the other. Functions in this form are called implicit functions.