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

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    Inverse Sin Square
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Discussion Overview

The discussion centers around finding the derivative of the function \( y = \sin^{-1}(x^2 + 2x) \). Participants explore various approaches to differentiate this expression, focusing on the application of the arcsine derivative formula and implicit differentiation techniques.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants emphasize the need for the original poster to share their attempts or thoughts on the problem to facilitate guidance.
  • One participant suggests using the derivative formula for arcsine, stating that if \( y = \arcsin(u(x)) \), then \( \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \).
  • Another participant proposes an implicit differentiation approach, starting from \( \sin(y) = u(x) \) and deriving \( \frac{dy}{dx} \) through implicit differentiation.
  • The same participant provides a step-by-step breakdown of the implicit differentiation process, ultimately leading to the same derivative formula for arcsine.

Areas of Agreement / Disagreement

Participants generally agree on the need for the original poster to provide more context about their understanding and attempts. However, there is no consensus on a specific method to solve the problem, as various approaches are suggested without resolution of which is preferable.

Contextual Notes

The discussion does not clarify specific assumptions regarding the domain of \( u(x) = x^2 + 2x \) or the conditions under which the derivative formula applies. There is also no resolution on the completeness of the proposed methods.

rahulk1
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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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