Tuesday, August 22, 2023

Laplace Transforms | Function and Transform (Examples)

You can find Laplace transforms below. We wish everyone good work.





You can find sample solutions below.





Sunday, August 20, 2023

General Rules of Differantiation

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In the following, u, v, w are functions of x; a, b, c, n are constants [restricted if indicated]; e=2.71828... is the natural base of logarithms; In u is the natural loraithm of u [i.e. the loraithm to the base e] where it is assumed thad u>0 and all angles are in radians.




Graphs of Hyperbolic Functions (Sinhx, Coshx, Tanhx, Cothx Graphics)

y=sinh x graph:




y=cosh x graph:




y=tanh x graph:




y=coth x graph:




y=sech x graph:




y=csch x graph:



Expression of Hyperbolic Functions In Terms of Others (Table)

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categories: 

 In the following we assume x>0.




Sunday, November 27, 2022

$sec^{-1} x^2$ Derivative | What is derivative of sec x^2?

Hello everyone. You can reach derivative of $sec^{-1} x^2$ in this lesson. We wish you good work.


sec -1 derivative


What is derivative of $sec^{-1} x^2$?


$Sec^{-1} u$ derivative formulas:

$\frac{d}{dx}sec^{-1} u=\frac{1}{|u|.\sqrt{u^2-1}}.\frac{du}{dx}$


Now let's answer our question.


Differentiate $y=sec^{-1} x^2$


Solution For $x^2$ > 1 > 0,

>>> $\frac{dy}{dx}=\frac{1}{|x|\sqrt{(x^2)^2-1}}.\frac{d}{dx}x^2$

>>> $=\frac{2x}{x^2\sqrt{x^4-1}}=\frac{2}{x.\sqrt{x^4-1}}$

Answer = $\frac{2}{x.\sqrt{x^4-1}}$

Tuesday, October 18, 2022

4-x^2 graph | What is the graph of $4-x^2$

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You can reach 4-x^2 graph in the below.

Graph of $4-x^2$


$y=4-x^2$ graphics:

A parabola that opens down is said to be “concave down”. The point (0, 4) is known as the vertex.

Coordinate Points :

x,y = (0,4)
x,y = (1,3)
x,y = (-1,3)
x,y = (2,0)
x,y = (-2,0)
x,y = (3,-5)
x,y = (-3,-5)


graph of 4-x^2


$4-x^2$ graphic




We wish everyone good work.

Friday, October 14, 2022

What is the slope of the tangent of f(x) at the point =1 ?



$f(x)=(x^2-4)^3$

What is the slope of the tangent of f(x) at the point=1?


Solution:

Derivative of f(x):

$f'(x)=[(x^2-4)^3]'$

$=3.(x^2-4)^2.(x^2-4)'$

$=3.(x^2-4)^2.2x$

$=6x.(x^2-4)^2$

At point x=1, the slope of the tangent of the function f is

$f'(1)=6.(1).(1^2-4)^2=6(1).(-3)^2=54$