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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List of Derivative Rules | All Derivative Rules

You can find all the derivative rules below. We wish everyone good work and good lessons...

Below is a list of all the derivative rules.

Constant Rule:

$f(x)=c$ then $f'(x)=0$

Constant Multiple Rule:

$g(x)=c.f(x)$ then $g'(x)=c.f'(x)$

Power Rule:

$f(x)=x^n$ then $f'(x)=nx^{n-1}$

Sum and Difference Rule:

$h(x)=f(x) ± g(x)$ then $h'(x)=f'(x) ± g'(x)$

Product Rule:

$h(x)=f(x).g(x)$ then $h'(x)=f'(x).g(x)+f(x).g'(x)$

Quotient Rule:

$h(x)=\frac{f(x)}{g(x)}$ then $h'(x)=\frac{f'(x).g(x)-f(x).g'(x)}{g(x)^2}$

Chain Rule:

$h(x)=f(g(x))$ then $h'(x)=f'(g(x)).g'(x)$

Trig Derivatives:

$f(x)=sin(x)$ then $f'(x)=cos(x)$

$f(x)=cos(x)$ then $f'(x)=-sin(x)$

$f(x)=tan(x)$ then $f'(x)=sec^2(x)$

$f(x)=sec(x)$ then $f'(x)=sec(x).tan(x)$

$f(x)=cot(x)$ then $f'(x)=-csc^2(x)$

$f(x)=csc(x)$ then $f'(x)=-csc(x).cot(x)$

Exponential Derivatives:

$f(x)=a^x$ then $f'(x)=ln(a).a^x$

$f(x)=e^x$ then $f'(x)=e^x$

$f(x)=a^{g(x)}$ then $f'(x)=ln(a).a^{g(x)}.g'(x)$

$f(x)=e^{g(x)}$ then $f'(x)=e^{g(x)}.g'(x)$

Logarithm Derivatives:

$f(x)=log_a(x)$ then $f'(x)=\frac{1}{ln(a).x}$

$f(x)=lnx$ then $f'(x)=\frac{1}{x}$

$f(x)=log_a(g(x))$ then $f'(x)=\frac{g'(x)}{ln(a).g(x)}$

$f(x)=ln(g(x))$ then $f'(x)=\frac{g'(x)}{g(x)}$

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arccot(x) integral | What is integrate of arccotx or cot^-1x?

Greetings dear friends. In this article, we will share with you what is the integrate of arccot(x).

arccot integrate

Integral of arccot(x) = $x.arccot(x)+\frac{1}{2}.ln(1+x^2)+C$

>>> $arccot(x)=cot^{-1}x$

>>> $\int arccot(x).dx=\int cot^{-1}x.dx$

>>> $\int arccot(x).dx=x.arccot(x)+\frac{1}{2}.ln(1+x^2)+C$

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