Differentiate tan (x¹3t) with respect to x, assuming that t is implicitly a function of x. dt (Use symbolic notation and fractions where needed. Use t' in place of dr.)
Differentiate tan (x¹3t) with respect to x, assuming that t is implicitly a function of x. dt (Use symbolic notation and fractions where needed. Use t' in place of dr.)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Differentiation of \( \tan \left( x^{\frac{1}{3}} t \right) \)
In this problem, we are asked to differentiate \( \tan \left( x^{\frac{1}{3}} t \right) \) with respect to \( x \), assuming that \( t \) is implicitly a function of \( x \).
**Given:**
Differentiate \( \tan \left( x^{\frac{1}{3}} t \right) \) with respect to \( x \), assuming that \( t \) is implicitly a function of \( x \).
(Note: Use symbolic notation and fractions where needed. Use \( t' \) in place of \( \frac{dt}{dx} \).)
**Solution:**
To solve this problem, we'll apply the chain rule and the product rule for differentiation.
1. **Chain Rule**:
The general form of the chain rule for differentiation states that if \( y = f(g(x)) \), then \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x). \]
2. **Product Rule**:
The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then \[ \frac{d}{dx} [u(x)v(x)] = u(x) v'(x) + v(x) u'(x). \]
Given: \( y = \tan \left( x^{\frac{1}{3}} t \right) \)
First, let \( u = x^{\frac{1}{3}} t \). Then, \( y = \tan(u) \).
Taking the derivative of \( y \) with respect to \( x \) we get:
\[ \frac{d}{dx} \left( \tan(u) \right) = \sec^2(u) \cdot \frac{du}{dx}. \]
Next, we need to find \( \frac{du}{dx} \):
\[ u = x^{\frac{1}{3}} t. \]
Let \( f(x) = x^{\frac{1}{3}} \) and \( g(x) = t \).
Applying the product rule:
\[
\frac{du}{dx} = \frac{d}{dx} \left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F575d721b-26eb-4db6-af93-50c207cd3fab%2F7099200f-b797-47ea-989e-37409eab1b03%2F7yfoi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Differentiation of \( \tan \left( x^{\frac{1}{3}} t \right) \)
In this problem, we are asked to differentiate \( \tan \left( x^{\frac{1}{3}} t \right) \) with respect to \( x \), assuming that \( t \) is implicitly a function of \( x \).
**Given:**
Differentiate \( \tan \left( x^{\frac{1}{3}} t \right) \) with respect to \( x \), assuming that \( t \) is implicitly a function of \( x \).
(Note: Use symbolic notation and fractions where needed. Use \( t' \) in place of \( \frac{dt}{dx} \).)
**Solution:**
To solve this problem, we'll apply the chain rule and the product rule for differentiation.
1. **Chain Rule**:
The general form of the chain rule for differentiation states that if \( y = f(g(x)) \), then \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x). \]
2. **Product Rule**:
The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then \[ \frac{d}{dx} [u(x)v(x)] = u(x) v'(x) + v(x) u'(x). \]
Given: \( y = \tan \left( x^{\frac{1}{3}} t \right) \)
First, let \( u = x^{\frac{1}{3}} t \). Then, \( y = \tan(u) \).
Taking the derivative of \( y \) with respect to \( x \) we get:
\[ \frac{d}{dx} \left( \tan(u) \right) = \sec^2(u) \cdot \frac{du}{dx}. \]
Next, we need to find \( \frac{du}{dx} \):
\[ u = x^{\frac{1}{3}} t. \]
Let \( f(x) = x^{\frac{1}{3}} \) and \( g(x) = t \).
Applying the product rule:
\[
\frac{du}{dx} = \frac{d}{dx} \left(
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