Problem 52
Question
In Exercises \(49-70\) , find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sin ^{-1}(1-t) $$
Step-by-Step Solution
Verified Answer
The derivative is \(-\frac{1}{\sqrt{2t-t^2}}\).
1Step 1: Identify the Function Type
The function given is **inverse sine**, specifically \( y = \sin^{-1}(1-t) \). This requires differentiation using the rules of inverse trigonometric functions.
2Step 2: Recall the Derivative Formula
The derivative of \( \sin^{-1}(x) \) with respect to \( x \) is given by \( \frac{d}{dx} \left[ \sin^{-1}(x) \right] = \frac{1}{\sqrt{1 - x^2}} \). We will use this formula to find the derivative of our function.
3Step 3: Apply the Chain Rule
Since our function is \( y = \sin^{-1}(1-t) \), we need to use the chain rule to differentiate it. Let \( u = 1-t \). Then \( y = \sin^{-1}(u) \). Find the derivative of \( u \) with respect to \( t \), \( \frac{du}{dt} = -1 \).
4Step 4: Differentiate the Outer Function
Differentiate \( y = \sin^{-1}(u) \) with respect to \( u \):\[ \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}}. \]
5Step 5: Use the Chain Rule to Complete Differentiation
Combine \( \frac{dy}{du} \) and \( \frac{du}{dt} \) using the chain rule: \[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = \frac{1}{\sqrt{1-(1-t)^2}} \cdot (-1) = -\frac{1}{\sqrt{1-(1-t)^2}}. \]
6Step 6: Simplify the Expression
Simplify the expression within the square root: \[ -\frac{1}{\sqrt{1 - (1 - 2t + t^2)}} = -\frac{1}{\sqrt{2t - t^2}}. \]
Key Concepts
Inverse Trigonometric FunctionsChain RuleCalculus Derivatives
Inverse Trigonometric Functions
Inverse trigonometric functions are essential in calculus when finding angles whose trigonometric ratios are known. These functions are often represented with a "-1" superscript, such as \( \sin^{-1}(x) \), which stands for the arc sine of \( x \). These functions are defined as the inverse of their respective trigonometric functions, making them unique and important in calculus.
When dealing with derivatives of inverse trigonometric functions, it's crucial to recall their specific derivative formulas. For the inverse sine function, \( \sin^{-1}(x) \), its derivative is \( \frac{1}{\sqrt{1 - x^2}} \). This formula is derived using implicit differentiation and the Pythagorean identity. Understanding how each inverse function is differentiated allows for effective problem solving when these functions appear complexly, as in a nested function context.
When dealing with derivatives of inverse trigonometric functions, it's crucial to recall their specific derivative formulas. For the inverse sine function, \( \sin^{-1}(x) \), its derivative is \( \frac{1}{\sqrt{1 - x^2}} \). This formula is derived using implicit differentiation and the Pythagorean identity. Understanding how each inverse function is differentiated allows for effective problem solving when these functions appear complexly, as in a nested function context.
Chain Rule
The chain rule is a vital tool in calculus used when differentiating composite functions. It allows you to differentiate a function that is composed of multiple nested functions. That's why it’s handy in our exercise, where we have \( y = \sin^{-1}(1-t) \).
To apply the chain rule, first identify the "inner function" and the "outer function." Here, the inner function is \( u = 1-t \), and the outer function is \( y = \sin^{-1}(u) \). The chain rule states that to differentiate \( y \) with respect to \( t \), you multiply the derivative of the outer function with respect to the inner function by the derivative of the inner function with respect to \( t \):
To apply the chain rule, first identify the "inner function" and the "outer function." Here, the inner function is \( u = 1-t \), and the outer function is \( y = \sin^{-1}(u) \). The chain rule states that to differentiate \( y \) with respect to \( t \), you multiply the derivative of the outer function with respect to the inner function by the derivative of the inner function with respect to \( t \):
- \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \)
Calculus Derivatives
Derivatives in calculus describe how a function changes as its input changes. They are foundational in understanding dynamic systems as they provide the rate of change. For instance, in our exercise, we are finding how \( y \) changes with \( t \) in the function \( y = \sin^{-1}(1-t) \).
Calculating derivatives often involves using rules like the power rule, product rule, quotient rule, or chain rule, depending on the complexity and structure of the function at hand.
Derivatives help solve a variety of problems, from finding tangent lines to optimization problems in industries. By understanding the derivative of an inverse trigonometric function through the chain rule, you combine their functional properties to effectively tackle a broader range of calculus problems, illustrating the sophistication yet logical beauty of calculus derivatives.
Calculating derivatives often involves using rules like the power rule, product rule, quotient rule, or chain rule, depending on the complexity and structure of the function at hand.
Derivatives help solve a variety of problems, from finding tangent lines to optimization problems in industries. By understanding the derivative of an inverse trigonometric function through the chain rule, you combine their functional properties to effectively tackle a broader range of calculus problems, illustrating the sophistication yet logical beauty of calculus derivatives.
Other exercises in this chapter
Problem 51
Evaluate the integrals. \(\int_{1}^{\sqrt{2}} x 2^{\left(x^{2}\right)} d x\)
View solution Problem 52
Evaluate the integrals in Exercises \(51-60 .\) $$ \int_{0}^{\ln 2} \tanh 2 x d x $$
View solution Problem 52
Evaluate the integrals in Exercises \(41-62\). $$ \int t^{3} e^{\left(t^{4}\right)} d t $$
View solution Problem 52
Evaluate the integrals in Exercises \(37-54\). $$ \int_{0}^{\pi / 12} 6 \tan 3 x d x $$
View solution