Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To find the first derivative of [tex]\( z = \tan(\sin^7(t)) \)[/tex] with respect to [tex]\( t \)[/tex], we will apply the chain rule of differentiation. Here’s a step-by-step approach:
1. Identify the outer function and its argument:
The outer function is [tex]\(\tan(x)\)[/tex], and its argument is [tex]\(x = \sin^7(t)\)[/tex].
2. Differentiate the outer function:
The derivative of [tex]\( \tan(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \sec^2(x) \)[/tex]. Therefore,
[tex]\[ \frac{d}{dx} \tan(x) = \sec^2(x). \][/tex]
3. Apply the outer function's derivative to its argument:
So, we get:
[tex]\[ \frac{d}{dx} \tan(\sin^7(t)) = \sec^2(\sin^7(t)). \][/tex]
4. Identify the inner function:
The inner function is [tex]\( x = \sin^7(t) \)[/tex].
5. Differentiate the inner function:
The inner function [tex]\( \sin^7(t) \)[/tex] is a composite function itself, and can be written as:
[tex]\[ (\sin(t))^7. \][/tex]
Using the chain rule again on this inner function:
- First, we differentiate power of [tex]\( \sin(t) \)[/tex], which gives us:
[tex]\[ 7(\sin(t))^6. \][/tex]
- Then, we multiply by the derivative of [tex]\( \sin(t) \)[/tex] itself:
[tex]\[ \frac{d}{dt} \sin(t) = \cos(t). \][/tex]
6. Combine these derivatives:
By combining the results from steps 3 and 5, we get:
[tex]\[ \frac{d}{dt} \tan(\sin^7(t)) = \sec^2(\sin^7(t)) \cdot 7 (\sin(t))^6 \cdot \cos(t). \][/tex]
7. Simplify the expression using [tex]\( \sec(x) = 1 / \cos(x) \)[/tex]:
We know that [tex]\(\sec(x) = 1 / \cos(x)\)[/tex], so:
[tex]\[ \sec^2(x) = (1 / \cos(x))^2 = \tan^2(x) + 1. \][/tex]
Thus, the first derivative of [tex]\( z = \tan(\sin^7(t)) \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \frac{dz}{dt} = 7 (\tan(\sin^7(t))^2 + 1) (\sin(t))^6 \cos(t). \][/tex]
So, the detailed solution opines that:
[tex]\[ \frac{dz}{dt} = 7 (\tan(\sin^7(t))^2 + 1) \sin^6(t) \cos(t). \][/tex]
1. Identify the outer function and its argument:
The outer function is [tex]\(\tan(x)\)[/tex], and its argument is [tex]\(x = \sin^7(t)\)[/tex].
2. Differentiate the outer function:
The derivative of [tex]\( \tan(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \sec^2(x) \)[/tex]. Therefore,
[tex]\[ \frac{d}{dx} \tan(x) = \sec^2(x). \][/tex]
3. Apply the outer function's derivative to its argument:
So, we get:
[tex]\[ \frac{d}{dx} \tan(\sin^7(t)) = \sec^2(\sin^7(t)). \][/tex]
4. Identify the inner function:
The inner function is [tex]\( x = \sin^7(t) \)[/tex].
5. Differentiate the inner function:
The inner function [tex]\( \sin^7(t) \)[/tex] is a composite function itself, and can be written as:
[tex]\[ (\sin(t))^7. \][/tex]
Using the chain rule again on this inner function:
- First, we differentiate power of [tex]\( \sin(t) \)[/tex], which gives us:
[tex]\[ 7(\sin(t))^6. \][/tex]
- Then, we multiply by the derivative of [tex]\( \sin(t) \)[/tex] itself:
[tex]\[ \frac{d}{dt} \sin(t) = \cos(t). \][/tex]
6. Combine these derivatives:
By combining the results from steps 3 and 5, we get:
[tex]\[ \frac{d}{dt} \tan(\sin^7(t)) = \sec^2(\sin^7(t)) \cdot 7 (\sin(t))^6 \cdot \cos(t). \][/tex]
7. Simplify the expression using [tex]\( \sec(x) = 1 / \cos(x) \)[/tex]:
We know that [tex]\(\sec(x) = 1 / \cos(x)\)[/tex], so:
[tex]\[ \sec^2(x) = (1 / \cos(x))^2 = \tan^2(x) + 1. \][/tex]
Thus, the first derivative of [tex]\( z = \tan(\sin^7(t)) \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[ \frac{dz}{dt} = 7 (\tan(\sin^7(t))^2 + 1) (\sin(t))^6 \cos(t). \][/tex]
So, the detailed solution opines that:
[tex]\[ \frac{dz}{dt} = 7 (\tan(\sin^7(t))^2 + 1) \sin^6(t) \cos(t). \][/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.