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

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    Inverse Sin Square
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SUMMARY

The discussion focuses on finding the derivative of the function \( y = \sin^{-1}(x^2 + 2x) \). The derivative is determined using the formula \( \frac{d}{dx}\arcsin(u(x)) = \frac{1}{\sqrt{1-u^2}}\,\frac{du}{dx} \). By applying implicit differentiation and substituting back for \( y \), the final expression for the derivative is \( \frac{dy}{dx} = \frac{1}{\sqrt{1 - (x^2 + 2x)^2}} \cdot (2x + 2) \). This method provides a clear pathway to calculate derivatives of inverse sine functions involving polynomial expressions.

PREREQUISITES
  • Understanding of inverse trigonometric functions, specifically arcsine.
  • Familiarity with implicit differentiation techniques.
  • Knowledge of basic calculus, including derivatives and chain rule.
  • Ability to manipulate algebraic expressions and functions.
NEXT STEPS
  • Study the application of the chain rule in calculus.
  • Learn more about implicit differentiation and its applications.
  • Explore the properties and graphs of inverse trigonometric functions.
  • Practice finding derivatives of composite functions involving polynomials and trigonometric identities.
USEFUL FOR

Students learning calculus, mathematics educators, and anyone seeking to understand the differentiation of inverse trigonometric functions involving polynomial expressions.

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if y = sin inverse (x square + 2x) find dy/dx
 
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rahulk said:
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 .
 
kaliprasad said:
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
 
rahulk said:
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:

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

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

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

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

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

Implicitly differentiate:

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

Hence:

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

Back-substitute for $y$:

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

And thus:

$$\d{y}{x}=\frac{1}{\sqrt{1-u^2}}\,\d{u}{x}$$

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

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