8 y' if y = log2 (√x+1).

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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I need help calculating the derivitve

The problem asks us to find \( y' \) if \( y = \log_2 \left( \frac{8}{\sqrt{4x+1}} \right) \).

To find the derivative \( y' \), you can use the chain rule and properties of logarithms. Let's step through it:

### Step 1: Simplify the expression
Using properties of logarithms, the given expression can be rewritten as:
\[ y = \log_2(8) - \log_2(\sqrt{4x+1}) \]
\[ y = 3 - \frac{1}{2} \log_2(4x+1) \]

### Step 2: Differentiate
We differentiate using the chain rule:
\[ y' = 0 - \frac{1}{2} \cdot \frac{1}{(4x+1) \ln(2)} \cdot 4 \]

### Conclusion
\[ y' = -\frac{2}{(4x+1) \ln(2)} \]

This expression represents the derivative of the original function. Use this derivative to analyze the behavior and rate of change of the function for various values of \( x \).
Transcribed Image Text:The problem asks us to find \( y' \) if \( y = \log_2 \left( \frac{8}{\sqrt{4x+1}} \right) \). To find the derivative \( y' \), you can use the chain rule and properties of logarithms. Let's step through it: ### Step 1: Simplify the expression Using properties of logarithms, the given expression can be rewritten as: \[ y = \log_2(8) - \log_2(\sqrt{4x+1}) \] \[ y = 3 - \frac{1}{2} \log_2(4x+1) \] ### Step 2: Differentiate We differentiate using the chain rule: \[ y' = 0 - \frac{1}{2} \cdot \frac{1}{(4x+1) \ln(2)} \cdot 4 \] ### Conclusion \[ y' = -\frac{2}{(4x+1) \ln(2)} \] This expression represents the derivative of the original function. Use this derivative to analyze the behavior and rate of change of the function for various values of \( x \).
Expert Solution
Step 1

Given the function y=log284x+1.

The objective is to find the derivative y'.

Formula used:

(1) Chain rule of differentiation is ddxf(g(x))=f'(g(x))g'(x), where fx and gx are differentiable function.

(2) ddxloga(x)=1(x ln a)

(3) Quotient rule of differentiation is ddxfxgx=gxf'x-fxg'xgx2. where fx and gx are differentiable function

(4) Power rule of differentiation is ddxxn=nxn-1, for all real n.

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