If y = sin inverse (x square + 2x) find dy/dx

[rahulk1]

if y = sin inverse (x square + 2x) find dy/dx

 

[kaliprasad]

if y = sin inverse (x square + 2x) find dy/dx

we are here to help you and not solve your problem, kindly inform what you have tried so that we can guide you .

 

[rahulk1]

we are here to help you and not solve your problem, kindly inform what you have tried so that we can guide you .

I just trying to learn mathematics that's why I am here if anybody can help please help

 

[MarkFL]

I just trying to learn mathematics that's why I am here if anybody can help please help

There are online calculators that will spit out the answer to this question, however if you want genuine help, we need to know what you've tried, or what your thoughts are on how to begin.

Suppose we are given:

\(\displaystyle y=\arcsin\left(u(x)\right)\)

And we are asked to find \(\displaystyle \d{y}{x}\). Now one approach would be to memorize the following formula:

\(\displaystyle \frac{d}{dx}\arcsin\left(u(x)\right)=\frac{1}{\sqrt{1-u^2}}\,\d{u}{x}\)

Another approach would be to derive the formula by rewriting the given equation as follows:

\(\displaystyle \sin(y)=u(x)\)

Implicitly differentiate:

\(\displaystyle \cos(y)\d{y}{x}=\d{u}{x}\)

Hence:

\(\displaystyle \d{y}{x}=\frac{1}{\cos(y)}\,\d{u}{x}\)

Back-substitute for $y$:

\(\displaystyle \d{y}{x}=\frac{1}{\cos\left(\arcsin\left(u(x)\right)\right)}\,\d{u}{x}\)

And thus:

\(\displaystyle \d{y}{x}=\frac{1}{\sqrt{1-u^2}}\,\d{u}{x}\)

So, can you now apply this formula to the given question?