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91Ó°ÊÓ

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\sin ^{-1}(1-t)$$

Short Answer

Expert verified
The derivative \( \frac{dy}{dt} \) is \(-\frac{1}{\sqrt{t(2-t)}}\).

Step by step solution

01

Define the Function

The function provided is \( y = \sin^{-1}(1-t) \). Our goal is to find \( \frac{dy}{dt} \), the derivative of \( y \) with respect to the variable \( t \).
02

Identify the Chain Rule Usage

The given function is a composition of functions \( y = \sin^{-1}(u) \), where \( u = 1-t \). We will use the chain rule, which states \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \).
03

Differentiate the Inverse Sine Function

The derivative of \( y = \sin^{-1}(u) \) with respect to \( u \) is \( \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \).
04

Differentiate the Inner Function

The derivative of \( u = 1-t \) with respect to \( t \) is \( \frac{du}{dt} = -1 \).
05

Apply the Chain Rule

Substitute the derivatives from Steps 3 and 4 into the chain rule expression: \[ \frac{dy}{dt} = \frac{1}{\sqrt{1-(1-t)^2}} \cdot (-1) \].
06

Simplify the Expression

Simplify the expression \( \frac{1}{\sqrt{1-(1-t)^2}} \cdot (-1) \). First, simplify the term inside the square root: \( 1-(1-t)^2 = t(2-t) \). Therefore, \( \frac{dy}{dt} = \frac{-1}{\sqrt{t(2-t)}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The Chain Rule is a foundational tool in calculus for finding the derivative of composite functions. A composite function is essentially a function within another function. Imagine having two functions, where one is nested inside the other. The Chain Rule tells us how to differentiate such functions by breaking them down into more manageable parts.
\[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \]
This formula expresses that if you have a function \( y = g(f(t)) \), you first find the derivative of \( y \) with respect to the inner function \( u = f(t) \) which gives you \( \frac{dy}{du} \). Next, you find the derivative of \( u \) with respect to \( t \), denoted \( \frac{du}{dt} \). Finally, multiply these derivatives together to get the derivative of \( y \) with respect to \( t \).

Using the Chain Rule simplifies the differentiation of complex expressions, making it indispensable, especially when multiplication would otherwise become cumbersome. Remember, the Chain Rule applies only when you can identify inner and outer functions acting on some variable.
Inverse Trigonometric Functions
Inverse Trigonometric Functions reverse the standard trigonometric functions to solve for angles. When you see a notation like \( \sin^{-1} \, \), it is referring not to a reciprocal, but to the inverse function. These functions, denoted as \( \sin^{-1} \, \), \( \cos^{-1} \, \), \( \tan^{-1} \) etc., allow us to determine an angle when given a ratio of sides of a triangle or another trigonometric expression.

In differentiation, inverse trigonometric functions require unique derivative rules due to their non-linear behavior. For instance, the derivative of \( y = \sin^{-1}(u) \) with respect to \( u \) is:
\[ \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \]
This expression reflects the geometric nature of the \( \sin^{-1} \, \) function, ensuring the output is real and within expected bounds. These derivatives can often involve complex algebra, especially when embedded in other expressions, but are central to solving many calculus problems. It's crucial to familiarize yourself with these derivatives to handle integration and differentiation tasks involving inverse trigonometric functions efficiently.
Differentiation Techniques
Differentiation Techniques refer to the variety of methods used to find the derivative of a function. Each technique is applicable under specific circumstances, depending on the form of the function involved. Here's a quick rundown of a few essential techniques you may encounter:
  • **The Chain Rule**: Used for differentiating composite functions, as explained earlier.
  • **Product Rule**: Used when differentiating a product of two functions; it states \( \frac{d}{dt}[u(t) \cdot v(t)] = u'(t)v(t) + u(t)v'(t) \).
  • **Quotient Rule**: Useful when dealing with a division of two functions; expressed as \( \frac{d}{dt}\left[\frac{u(t)}{v(t)}\right] = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \).
  • **Inverse Trigonometric Derivatives**: As discussed, derivatives of functions like \( \sin^{-1} \), \( \cos^{-1} \), etc.
  • **Implicit Differentiation**: Sometimes functions are given implicitly rather than explicitly, requiring differentiation strategies tailored for such expressions.
This exercise regarding \( y = \sin^{-1}(1-t) \) primarily calls for the Chain Rule, demonstrating how multiple techniques can sometimes be nested together. Practicing these techniques with a variety of functions is the best way to increase proficiency.

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