To solve the problem of finding $\frac{dy}{dt}$ given the function $y = \ln \left( \frac{t^2 + 9}{\sqrt{8 - t}} \right)$, you will need to apply the chain rule and quotient rule from calculus.### Solution Steps:1. **Identify Parts of the Function:** - Let $u = t^2 + 9$ and $v = \sqrt{8 - t}$. - The function becomes $y = \ln \left( \frac{u}{v} \right)$.2. **Apply Logarithmic Differentiation:** - The derivative of $\ln \left( \frac{u}{v} \right)$ is $\frac{1}{\left( \frac{u}{v} \right)} \cdot \left( \frac{du/dt \cdot v - u \cdot dv/dt}{v^2} \right)$.3. **Find Derivatives of $u$ and $v$:** - $du/dt = 2t$. - $dv/dt$ requires implicit differentiation: - $v = (8-t)^{1/2}$, so $dv/dt = -\frac{1}{2}(8-t)^{-1/2}$.4. **Substitute into the Derivatives:** - Compute the derivative $\frac{du}{v^2} = \frac{2t}{(8-t)}$. - Compute the second part: $- \frac{u \cdot \frac{1}{2}(8-t)^{-3/2}}{v^2} = - \frac{(t^2 + 9)}{2(8-t)^{-3/2}}$.By solving these steps, you'll find the solution for $\frac{dy}{dt}$. The crucial aspect is correctly applying derivative rules and simplifying the expressions. Once calculated, substitute the values back to get the explicit derivative $\frac{dy}{dt} = \text{[your solution]}$. **Problem Statement:**Find $\frac{dy}{dx}$ if $ y = \ln(x^3 \sqrt{x+1}) $.**Solution:**\[\frac{dy}{dx} = \boxed{\phantom{x}}\]**Explanation:**To solve this problem, we need to differentiate the given function $ y = \ln(x^3 \sqrt{x+1}) $ with respect to $ x $.### Steps for Differentiation:1. **Use Logarithmic Properties:** Simplify the expression inside the logarithm using properties of logarithms: \[ y = \ln(x^3 \cdot (x+1)^{1/2}) = \ln(x^3) + \ln((x+1)^{1/2}) \] This can be further simplified to: \[ y = 3\ln(x) + \frac{1}{2}\ln(x+1) \]2. **Differentiate:** Apply the derivative: - The derivative of $ 3\ln(x) $ is $\frac{3}{x}$. - The derivative of $ \frac{1}{2}\ln(x+1) $ is $\frac{1}{2} \cdot \frac{1}{x+1}$. Combine the derivatives: \[ \frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)} \]This is the derivative of the function. Fill in the box with the expression obtained above.

Given equation is y=lnt2+98-t. Compute the derivative with respect to t.…

Answered: t2 + 9 V8 - t dy if y = In dt Find dy…

Math

Calculus

t2 + 9 V8 - t dy if y = In dt Find dy dt

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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Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Question

To solve the problem of finding $\frac{dy}{dt}$ given the function $y = \ln \left( \frac{t^2 + 9}{\sqrt{8 - t}} \right)$, you will need to apply the chain rule and quotient rule from calculus.

### Solution Steps:

1. **Identify Parts of the Function:**
- Let $u = t^2 + 9$ and $v = \sqrt{8 - t}$.
- The function becomes $y = \ln \left( \frac{u}{v} \right)$.

2. **Apply Logarithmic Differentiation:**
- The derivative of $\ln \left( \frac{u}{v} \right)$ is $\frac{1}{\left( \frac{u}{v} \right)} \cdot \left( \frac{du/dt \cdot v - u \cdot dv/dt}{v^2} \right)$.

3. **Find Derivatives of $u$ and $v$:**
- $du/dt = 2t$.
- $dv/dt$ requires implicit differentiation:
- $v = (8-t)^{1/2}$, so $dv/dt = -\frac{1}{2}(8-t)^{-1/2}$.

4. **Substitute into the Derivatives:**
- Compute the derivative $\frac{du}{v^2} = \frac{2t}{(8-t)}$.
- Compute the second part: $- \frac{u \cdot \frac{1}{2}(8-t)^{-3/2}}{v^2} = - \frac{(t^2 + 9)}{2(8-t)^{-3/2}}$.

By solving these steps, you'll find the solution for $\frac{dy}{dt}$. The crucial aspect is correctly applying derivative rules and simplifying the expressions. Once calculated, substitute the values back to get the explicit derivative $\frac{dy}{dt} = \text{[your solution]}$.

**Problem Statement:**

Find $\frac{dy}{dx}$ if $ y = \ln(x^3 \sqrt{x+1}) $.

**Solution:**

\[
\frac{dy}{dx} = \boxed{\phantom{x}}
\]

**Explanation:**

To solve this problem, we need to differentiate the given function $ y = \ln(x^3 \sqrt{x+1}) $ with respect to $ x $.

### Steps for Differentiation:

1. **Use Logarithmic Properties:**

Simplify the expression inside the logarithm using properties of logarithms:

\[
y = \ln(x^3 \cdot (x+1)^{1/2}) = \ln(x^3) + \ln((x+1)^{1/2})
\]

This can be further simplified to:

\[
y = 3\ln(x) + \frac{1}{2}\ln(x+1)
\]

2. **Differentiate:**

Apply the derivative:

- The derivative of $ 3\ln(x) $ is $\frac{3}{x}$.

- The derivative of $ \frac{1}{2}\ln(x+1) $ is $\frac{1}{2} \cdot \frac{1}{x+1}$.

Combine the derivatives:

\[
\frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)}
\]

This is the derivative of the function. Fill in the box with the expression obtained above.

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