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Magazine Chapter 3: Problem 53
In Exercises \(41-58\) find \(d y / d t\) $$ y=\sqrt{1+\cos \left(t^{2}\right)} $$
Short Answer
Expert verified
\(\frac{dy}{dt} = \frac{-t \sin(t^2)}{\sqrt{1 + \cos(t^2)}}\)
Step by step solution
01
Identify the Function and Derivative Purpose
The function given is \(y = \sqrt{1 + \cos(t^2)}\). We need to find \(\frac{dy}{dt}\), the derivative of \(y\) with respect to \(t\).
02
Apply Chain Rule
Recognize this as a composite function and apply the chain rule. The outer function is the square root and the inner function is \(1 + \cos(t^2)\). The chain rule states \(\frac{d}{dt}f(g(t)) = f'(g(t)) \cdot g'(t)\).
03
Differentiate the Outer Function
The derivative of \(\sqrt{x}\) with respect to \(x\) is \(\frac{1}{2\sqrt{x}}\). Therefore, differentiating the outer function, we get \(\frac{1}{2\sqrt{1 + \cos(t^2)}}\).
04
Differentiate the Inner Function
Differentiate \(1 + \cos(t^2)\) using the chain rule again, where \(g(t) = \cos(t^2)\). The derivative of \(\cos(u)\) is \(-\sin(u)\), so \(-\sin(t^2)\), and the derivative of \(t^2\) with respect to \(t\) is \(2t\). Thus, \(\frac{d}{dt} \cos(t^2) = -2t\sin(t^2)\).
05
Combine Results Using the Chain Rule
By the chain rule, the derivative \(\frac{dy}{dt}\) is the product of the derivatives from Step 3 and Step 4. Thus:\[\frac{dy}{dt} = \frac{1}{2\sqrt{1 + \cos(t^2)}} \cdot (-2t \sin(t^2)).\]
06
Simplify the Result
Simplify the expression:\[\frac{dy}{dt} = \frac{-2t \sin(t^2)}{2\sqrt{1 + \cos(t^2)}} = \frac{-t \sin(t^2)}{\sqrt{1 + \cos(t^2)}}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
A derivative represents the rate at which a function is changing at any given point. In simple terms, it's how fast one quantity is changing in relation to another. This concept is vital in calculus because it helps us understand the behavior and slope of curves. When you find the derivative of a function, you determine its instantaneous rate of change.
- Consider the function given: \( y = \sqrt{1 + \cos(t^2)} \). You need to find \( \frac{dy}{dt} \), meaning you want to know how \( y \) changes as \( t \) changes.
- Derivatives are typically denoted by \( f'(x) \) or \( \frac{dy}{dx} \), where \( y = f(x) \).
Composite Function
Composite functions occur when one function is nested inside another function, essentially creating a function within a function. The notation often used is \( f(g(x)) \), where \( g(x) \) is an inner function, and \( f(x) \) is an outer function. Understanding this concept is crucial when dealing with problems where functions are layered.
- In the exercise, the composite function is \( y = \sqrt{1 + \cos(t^2)} \). The square root covers the entire expression \( 1 + \cos(t^2) \), making it the outer function.
- The expression \( 1 + \cos(t^2) \) becomes the inner function \( g(t) \). Here, \( \cos(t^2) \) adds another layer, with \( t^2 \) being affected directly by \( t \).
Trigonometric Differentiation
Trigonometric differentiation involves taking derivatives of trigonometric functions such as sine, cosine, tangent, etc. Understanding how these derivatives behave is vital when they are part of a more complex expression. Trig functions have specific rules that apply when differentiating them.For example:
- The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \).
- In our exercise, you differentiate \( \cos(t^2) \) to find \( g'(t) \). Here, we see \( g(t) = \cos(t^2) \), followed by \( u = t^2 \), and we use \( -2t\sin(t^2) \) for the derivative \( -\sin(t^2) \times 2t \).
