General Rules of Differantiation

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.

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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:

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Expression of Hyperbolic Functions In Terms of Others (Table)

 In the following we assume x>0.

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

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

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.

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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

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

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Derivative of $(x^2-4)^3$ | Find the derivative of the function $f(x)=(x^2-4)^3$

Find the derivative of the function $f(x)=(x^2-4)^3$

Solution

We need to derive the composite function $u^3$, where $u=x^2-4$. Consequently, we need to use the chain derivative.

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

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

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

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

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