1. Successive differentiation
-
1) \( y = 2x + \frac{4}{x} \), prove that \( x^2
\frac{d^2y}{dx^2} - y = 0 \)
-
2) \( y = e^{ax} \), show that \( (1 - x^2) y_2 - xy_1 = a^2 y
\)
-
3) \( y = \sec(\sec x) \), show that \( y_2 = y(2y^2 - 1) \)
-
4) \( y = Ae^{mx} + Be^{-mx}\), show that \( y_2 = -m^2y\)
- 5) \( y = Ae^x + Be^{-x}\), show that \( y_2 - y = 0\)
-
6) \( y = x^4 + 4x^3 + 10 \), then find \( y_2 \) and \( y_3 \)
-
7) \( y = 8x^5 - 4x^3 + 14x^2 \), then find \( y_2 \) and \( y_3
\) and \( y_4 \)
2. Partial derivative(click me)
You just need to watch a few minutes, its easy
-> Partial Derivative is such a differentiation where we derive
with respect to one variable, while keeping the other constant.
example_1:
f(x, y) = 9x3 + 5x2y + 9xy2
+ 3y2
δf / δx = 27x2 + 10xy + 9y2
δf / δy = 5x2
+ 18xy + 6y
2:
f(x, y) = 4x3y + 6x2 + 3y2x
+ 5y
δf / δx = 12x2y + 12x + 3y2
δf / δy = 4x3 + 6xy + 5
3:
f(x, y) = 7x2y3 + 5xy + 2y4
δf / δx = 14xy3 + 5y
δf / δy = 21x2y2 + 5x + 8y3
4:
f(x, y) = 5x4 + 2x2y2 +
y3
δf / δx = 20x3 + 4xy2
δf / δy = 4x2y + 3y2
5:
f(x, y) = 6x2 + 3xy + 2y2
δf / δx = 12x + 3y
δf / δy = 3x + 4y
6:
f(x, y) = x3y2 + 2xy + y
δf / δx = 3x2y2 + 2y
δf / δy = 2x3y + 2x + 1
3. Multiple integration(click me)
4. Integration(click me)
- \( \int dx = x + C \)
-
\( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \)
- \( \int \frac{1}{x} \, dx = \ln|x| + C \)
- \( \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} + C \)
- \( \int e^{mx} \, dx = \frac{1}{m} e^{mx} + C \)
- \( \int a^x \, dx = \frac{a^x}{\ln a} + C \)
- \( \int \cos x \, dx = \sin x + C \)
- \( \int \sin x \, dx = -\cos x + C \)
- \( \int \sec^2 x \, dx = \tan x + C \)
- \( \int \csc^2 x \, dx = -\cot x + C \)
- \( \int \sec x \tan x \, dx = \sec x + C \)
- \( \int \csc x \cot x \, dx = -\csc x + C \)
- \( \int \tan x \, dx = \ln|\sec x| + C \)
- \( \int \cot x \, dx = \ln|\sin x| + C \)
- \( \int \sec x \, dx = \ln|\sec x + \tan x| + C \)
- \( \int \csc x \, dx = \ln|\csc x - \cot x| + C \)
-
\( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(
\frac{x}{a} \right) + C \)
-
\( \int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln\left| \frac{a +
x}{a - x} \right| + C \)
-
\( \int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln\left| \frac{x -
a}{x + a} \right| + C \)
- \( \int 5x^3 \, dx \)
- \( \int \frac{1}{x} \, dx \)
- \( \int \frac{x^4 + 1}{x^2} \, dx \)
- \( \int \frac{1 + x}{x} \, dx \)
- \( \int (x^2 + e^x + 2^x) \, dx \)
- \( \int (4x^3 + 3x^2 - 2x + 5) \, dx \)
- \( \int (1 - 3x)(1 + x) \, dx \)
- \( \int (3x^{-1} + 4x^2 - 3x + 8) \, dx \)
- \( \int (8e^x - 4a^x + 3x^{-1} + \sqrt[4]{x}) \, dx \)
- \( \int (2x + 9)^5 \, dx \)
- \( \int (\sqrt{5x + 7})^3 \, dx \)
- \( \int ((x^3 + 2)^3 - 3x^2) \, dx \)
- \( \int \frac{dx}{x \ln x \ln(\ln x)} \)
5. Area finding(click me)
6. Rolles and mean theorem(click me)
Rolles and Means youtube
- Rolles
-
1. Solve for f(a)=f(b), if equal, move on.
2. Check if it is continuous on the closed intervals
[a,b].
-- If polynomial, continuous on (-∞, ∞)
-- If modulus (absolute value), continuous everywhere
-- If rational, denominator must not be equal to 0
-- If exponential (e^x), continuous everywhere
-- If roots, the inside of the radical must be ≥ 0
-- If log(x), argument must be > 0
3. Check if it is differentiable on the open intervals
(a,b).
-- If polynomial, differentiable on (-∞, ∞)
-- If modulus (absolute value), not differentiable where inside
= 0
-- If exponential (e^x), differentiable everywhere
-- If rational, differentiable where denominator ≠ 0
-- If roots, differentiable where inside > 0
-- If log(x), differentiable where x > 0
4. Find f'(x), change x to c, f '(c) = 0, solve for c, if
c is inside the given range, thats the answer.
- f(x) = x(x-3)
- \( f(x) = -3x \sqrt{x+1} \)
- f(x) = x2-5x+4, [1,4]
- f(x) = x2/3-1, [-8,8]
-
f(x) =
x2-2x-3⁄x+2, [-1,3]
-
f(x) = x2-1⁄x, [-1,1]
- f(x) = cos x, [0, 2π]
- Means
- Must follow, both step 2 and 3.
- f`(c) = f(b) - f(a) ⁄ b-a
- f(x) = x3, [0, 1]
- f(x) = x4-8x, [0, 2]
- \( f(x) = \frac{x+1}{x} \) , \( [-1 , 2] \)
- f(x) = sin x, [0, π]
7. Beta and gamma function(click me)
Youtube learning link
- Beta
-
Relation between Beta and Gamma functions: \(β(m, n) =
\dfrac{\Gamma(m) \Gamma(n)}{\Gamma(m + n)}\)
-
\(β(m, n) = \int_0^1 x^{m-1} (1 - x)^{n-1} \, dx \quad (m, n >
0)\)
- Symmetric property: \(β(m, n) = β(n, m)\)
-
\(β(m, n) = 2 \int_0^{\pi/2} (\sin \theta)^{2m-1} (\cos
\theta)^{2n-1} \, d\theta\)
- Gamma
- 1) \(\Gamma(n) = \int_0^{\infty} e^{-x} x^{n-1} \, dx\)
- 2) \(\Gamma(1) = 1\)
- 3) \(\Gamma(n + 1) = n \Gamma(n)\)
- 4) \(\Gamma(n + 1) = n!\)
- 5) \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\)
- 6) \(\Gamma(m) \Gamma(1 - m) = \dfrac{\pi}{\sin(\pi m)}\)
Exercise:
- 1) 1) Prove that, \(\Gamma(1 / 2) = \sqrt{\pi}\)
- 2) Evaluate \(B(3, 2)\)
-
3) Show that \(B(m, n) = \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m +
n)}\)
-
4) Find the value of \(B\left(\dfrac{1}{2},
\dfrac{1}{2}\right)\)
- 5) Prove that \(B(1, n) = \dfrac{1}{n}\)
- 6) Evaluate \(B(2, 3)\) using Gamma functions
-
7) Express \(B(m, n)\) as an integral involving sine and
cosine:
\(B(m, n) = 2 \int_0^{\pi/2} (\sin \theta)^{2m-1} (\cos
\theta)^{2n-1} \, d\theta\)
-
8) Evaluate \(\int_0^{\pi/2} \sin^4\theta \cos^2\theta \,
d\theta\) using Beta function
- 9) Evaluate \(\int_0^{\pi/2} \sin^5\theta \, d\theta\)
-
10) Show that \(\int_0^{\pi/2} \sin^{m-1}\theta
\cos^{n-1}\theta \, d\theta = \dfrac{1}{2}B\left(\dfrac{m}{2},
\dfrac{n}{2}\right)\)
-
11) If \(\Gamma(m)\Gamma(1 - m) = \dfrac{\pi}{\sin(\pi m)}\),
find the value of
\(\Gamma\left(\dfrac{3}{4}\right)\Gamma\left(\dfrac{1}{4}\right)\)
- 12) Evaluate \(\Gamma(5)\)
-
13) Prove that \(\Gamma(n + 1) = n\Gamma(n)\) for all positive
integers \(n\)
- 14) Show that \(\Gamma(n + 1) = n!\)
-
15) Find the value of \(\Gamma\left(\dfrac{3}{2}\right)\)
-
16) Prove that \(\Gamma\left(\dfrac{5}{2}\right) =
\dfrac{3}{4}\sqrt{\pi}\)
8. Differentiation, monotonicity, concavity(click me)
Youtube learning link
Step 1: Find \( f'(x) = 0 \), find the values of x, draw a
number line, point out the numbers, from -infinity, numbers,
+infinity, note down the maximum and minimum points. Put the
critical points in \( f(x) = 0 \) to find the y-axis values.
Step 2: Find \( f''(x) \) = 0, determine the concavity of
the function. Draw a number line, and note down the ups (+) and
downs (-). Put the fluctuation points in \( f(x) = 0 \) to find
the y-axis values.
Step 3:Plot the critical points and the fluctuation points,
(x,y) in graph, draw the graph.
- 1) \( f(x) = 4x + \frac{64}{x} \) at \( x = 3\)
- 2) \( f(x) = x^3 - 15x^2 + 75x \) at \( x = -8\)
- 3) \( f(x) = x^2 - 4x + 3 \) , find intervals
- 4) \( f(x) = x^3\) , find intervals
- 5) \( f(x) = 3x^4 + 4x^3 - 12x^2 + 2\) , find intervals
9.Absolute maxima and minima(click me)
Youtube learning link
Extreme value theorem:
If a function f is continuous on a finite closed interval [a, b],
then f has both a maximum and a minimum value on that
interval[a,b], then f has both an absolute minima and an absolute
maxima.
Shortcut to check if interval gap is huge.
For a continuous function f(x) on a closed interval [a,b], the
Extreme Value Theorem says: Absolute max and min occur either at
critical points or at the endpoints.
Critical points: the
points in f`(x) = 0.
Endpoints: given interval, [a,b]
- \( f(x) = 2x^3 - 15x^2 + 36x, \quad [1,5] \)
-
\( f(x) = 6\frac{x^4}{3} - 3\frac{x^1}{3}, \quad [-1,1] \)
- \( f(x) = 4x^2 - 12x + 10, \quad [1,2] \)
- \( f(x) = 8x - x^2 \)
- \( f(x) = \frac{3x}{\sqrt{4x^2 + 1}} \)
- \( f(x) = 1 + |9 - x^2|, \quad [-5,1] \)
- \( f(x) = |6 - 4x|, \quad [-3,3] \)