arcsin(x) integral | What is integrate of arcsinx or sin^-1x?

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

arcsin integrate

Integral of arcsin(x) = $x.arcsin(x)+\sqrt{1-x^2}+C$

>>> $arcsin(x)=sin^{-1}x$

>>> $\int arcsin(x).dx=\int sin^{-1}x.dx$

>>> $\int arcsin(x).dx=x.arcsin(x)+\sqrt{1-x^2}+C$

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$\int \frac{x}{x-1}dx$ integral | What is the integrate of x/x-1 ?

Hello everyone. This lesson we will tell the integrate of $\int \frac{x}{x-1}dx$.

What is the integral of x/x-1?

integral of x/x-1

First we separate the expression.

So;

$\frac{x}{x-1}=1+\frac{1}{x-1}$

So our answer;

$\int \frac{x}{x-1}dx=\int 1 dx+ \int \frac{1}{x-1}dx=x+ln|x-1|+C$

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$\sqrt{1+\sqrt{1+\sqrt{x}}}$ derivative | What is the derivative of $\frac{d}{dx}\sqrt{1+\sqrt{1+\sqrt{x}}}$?

Hello everyone. This lesson we will tell the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$.

What is the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$?

Compute the derivative of $\sqrt{1+\sqrt{1+\sqrt{x}}}$.

Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost “layer” we have the function $g(x)=1+\sqrt{1+\sqrt{x}}$ plugged into $f(x)=\sqrt{x}$, so applying the chain rule once gives

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

Not we need the derivative of $\sqrt{1+\sqrt{x}}$. Using the chain rule again:

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

So the original derivative is

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

=$\frac{1}{8\sqrt{x}.\sqrt{1+\sqrt{x}}.\sqrt{1+\sqrt{1+\sqrt{x}}}}$

Using the chain rule, the power rule, and the product rule, it is possible to avıid using the quotient rule entirely.

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Product Rule for Derivative | How to take the multiplication derivative?

Hello everyone. This lesson we will tell the product rule for derivative.

Product Rule for Derivative

product rule of derivative

Let f(x) and g(x) be two functions. Then the derivate of the product

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

We must follow this rule religiously and not succumb to the temptation of writing $(f(x)g(x))'=f'(x)g'(x)$ ; a faulty statement.

Example:

$(x^3.e^x)'=(x^3)'.e^x+x^3(e^x)'$

=$3x^2e^x+x^3e^x$

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Quotient Rule for Derivative | How to take the derivative of the quotient?

Hello everyone. This lesson we will tell the quotient rule for derivative.

Quotient Rule for Derivative

quotient rule derivative

Let ݂f(x) and ݃g(x) be two functions. Then the derivative of the quotient:

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

This is how the derivative of the quotient is taken. Now let's reinforce the issue with an example.

Example:

What is the derivative of $(\frac{x^3}{e^x})'$

$(\frac{x^3}{e^x})'=\frac{(x^3)'.e^x-x^3.(e^x)'}{(e^x)^2}$

$=\frac{3x^2e^x-x^3e^x}{(e^x)^2}$

$=\frac{x^2e^2.(3-x)}{(e^x)^2}$

$=\frac{x^2(3-x)}{e^x}$

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Graph of cotangent | What is the cot(x) graph drawing?

Hello everyone dear friends. In this lesson, we will share with you what the cotangent(x) graph is.

What is the graph of cotx?

The plot of cot(x) is as follows:

cotangent graph

graph of cot(x)

Both graphs are graphs of cotangent(x). You can use both. Thank you so much. We wish you success in your studies.

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$e^{1-x}dx$ integral | What is integral of e^1-x ?

You can reach integrate of e^1-x solution and answer.

What is integral of $e^{1-x}dx$ ?

integrate of e^1-x

Answer:

$\int e^{1-x}dx=?$

$\int e^{1-x}dx=-e^{1-x}+C$

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