Question
upstudy study bank question image url

\( 1 \leftarrow \) Find \( z^{\prime}(t) \), where \( z=\frac{1}{x}+\frac{1}{y}, x=t^{2}+2 t \), and \( y=t^{3}-3 \), in the following ways. a. Replace \( x \) and \( y \) to write \( z \) as a function of \( t \) and differentiate. b. Use the Chain Rule. a. Write \( z \) as a function of \( t \). \( z(t)=\frac{1}{t^{2}+2 t}+\frac{1}{t^{3}-3} \) b. To find the derivative using the Chain Rule, begin by finding the intermediate derivatives. \( \frac{\partial z}{\partial x}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)

Ask by Sullivan Williams. in the United States
Mar 17,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

To find \( z^{\prime}(t) \), where \( z = \frac{1}{x} + \frac{1}{y} \), \( x = t^2 + 2t \), and \( y = t^3 - 3 \), we can differentiate \( z \) with respect to \( t \) using two methods: **a. Direct Differentiation:** \[ z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] \[ z'(t) = -\frac{2t + 2}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} \] **b. Using the Chain Rule:** \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] \[ \frac{dx}{dt} = 2t + 2, \quad \frac{dy}{dt} = 3t^2 \] \[ z'(t) = -\frac{1}{x^2} (2t + 2) - \frac{1}{y^2} (3t^2) \] Thus, \( z^{\prime}(t) \) is: \[ z^{\prime}(t) = -\frac{2t + 2}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} \]

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Mind Expander

To differentiate \( z \) with respect to \( x \) using \( z=\frac{1}{x}+\frac{1}{y} \), we can apply the derivative rules: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] And since \( y \) is a function of \( t \), we also need to include the derivative of \( y \) with respect to \( x \) when applying the Chain Rule, which is often computed as \( \frac{\partial z}{\partial y} \cdot \frac{dy}{dx} \): \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] Now, using the Chain Rule, you can relate the derivatives of \( z \) to the derivatives of \( t \): To get \( z'(t) \) using the Chain Rule, you’ll need: 1. \( \frac{dx}{dt} = 2t + 2 \) 2. \( \frac{dy}{dt} = 3t^2 \) Combined, the Chain Rule gives: \[ z'(t) = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \] Inserting the expressions we found: \[ z'(t) = \left(-\frac{1}{x^2}\right)(2t+2) + \left(-\frac{1}{y^2}\right)(3t^2) \] This leads to the complete form for \( z'(t) \): Combine all components to derive the expression for \( z'(t) \).

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy