Solved on Feb 21, 2024

Find the derivative of the function $g(u) = \frac{u+1}{2u-1}$ using the definition of derivative. State the domain of $g(u)$ and its derivative in interval notation.

STEP 1

Assumptions
1. The function given is $g(u) = \frac{u+1}{2u-1}$ .
2. We will use the definition of the derivative, which is given by: $g'(u) = \lim_{h \to 0} \frac{g(u+h) - g(u)}{h}$
3. We will find the domain of the function and its derivative using interval notation.

STEP 2

First, we need to find the expression for $g(u+h)$ by substituting $u+h$ into the function $g(u)$ .
$g(u+h) = \frac{(u+h)+1}{2(u+h)-1}$

STEP 3

Simplify the expression for $g(u+h)$ .
$g(u+h) = \frac{u+h+1}{2u+2h-1}$

STEP 4

Now, we will set up the difference quotient for the definition of the derivative.
$g'(u) = \lim_{h \to 0} \frac{\frac{u+h+1}{2u+2h-1} - \frac{u+1}{2u-1}}{h}$

STEP 5

Find a common denominator for the terms in the numerator of the difference quotient.
$g'(u) = \lim_{h \to 0} \frac{(u+h+1)(2u-1) - (u+1)(2u+2h-1)}{h(2u+2h-1)(2u-1)}$

STEP 6

Expand the numerator of the difference quotient.
$g'(u) = \lim_{h \to 0} \frac{(2u^2 - u + 2uh - h - 1) - (2u^2 + u + 2uh + 2h^2 - 1)}{h(2u+2h-1)(2u-1)}$

STEP 7

Simplify the numerator by combining like terms.
$g'(u) = \lim_{h \to 0} \frac{-2u - 2h^2}{h(2u+2h-1)(2u-1)}$

STEP 8

Cancel out the $h$ in the numerator with the $h$ in the denominator.
$g'(u) = \lim_{h \to 0} \frac{-2u - 2h}{(2u+2h-1)(2u-1)}$

STEP 9

Now take the limit as $h$ approaches 0.
$g'(u) = \frac{-2u}{(2u-1)^2}$

STEP 10

The derivative of the function $g(u)$ is:
$g'(u) = \frac{-2u}{(2u-1)^2}$

STEP 11

To find the domain of the function $g(u)$ , we need to determine where the function is defined. The function is undefined where the denominator is zero.
$2u - 1 = 0$

STEP 12

Solve for $u$ to find the value that is not in the domain of $g(u)$ .
$u = \frac{1}{2}$

STEP 13

The domain of $g(u)$ is all real numbers except $u = \frac{1}{2}$ .
$\text{Domain of } g(u): (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

STEP 14

To find the domain of the derivative $g'(u)$ , we look at the denominator of the derivative.
$(2u-1)^2 = 0$

STEP 15

Solve for $u$ to find the value that is not in the domain of $g'(u)$ .
$u = \frac{1}{2}$

STEP 16

The domain of $g'(u)$ is the same as the domain of $g(u)$ because the derivative also has a denominator that cannot be zero.
$\text{Domain of } g'(u): (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$
The derivative of the function $g(u)$ is $g'(u) = \frac{-2u}{(2u-1)^2}$ , and the domain of both the function and its derivative is $(-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$ .

Was this helpful?

Find the derivative of the function $g(u) = \frac{u+1}{2u-1}$ using the definition of derivative. State the domain of $g(u)$ and its derivative in interval notation.

STEP 1

Assumptions
1. The function given is $g(u) = \frac{u+1}{2u-1}$ .
2. We will use the definition of the derivative, which is given by: $g'(u) = \lim_{h \to 0} \frac{g(u+h) - g(u)}{h}$
3. We will find the domain of the function and its derivative using interval notation.

STEP 2

First, we need to find the expression for $g(u+h)$ by substituting $u+h$ into the function $g(u)$ .
$g(u+h) = \frac{(u+h)+1}{2(u+h)-1}$

STEP 3

Simplify the expression for $g(u+h)$ .
$g(u+h) = \frac{u+h+1}{2u+2h-1}$

STEP 4

Now, we will set up the difference quotient for the definition of the derivative.
$g'(u) = \lim_{h \to 0} \frac{\frac{u+h+1}{2u+2h-1} - \frac{u+1}{2u-1}}{h}$

STEP 5

Find a common denominator for the terms in the numerator of the difference quotient.
$g'(u) = \lim_{h \to 0} \frac{(u+h+1)(2u-1) - (u+1)(2u+2h-1)}{h(2u+2h-1)(2u-1)}$

STEP 6

Expand the numerator of the difference quotient.
$g'(u) = \lim_{h \to 0} \frac{(2u^2 - u + 2uh - h - 1) - (2u^2 + u + 2uh + 2h^2 - 1)}{h(2u+2h-1)(2u-1)}$

STEP 7

Simplify the numerator by combining like terms.
$g'(u) = \lim_{h \to 0} \frac{-2u - 2h^2}{h(2u+2h-1)(2u-1)}$

STEP 8

Cancel out the $h$ in the numerator with the $h$ in the denominator.
$g'(u) = \lim_{h \to 0} \frac{-2u - 2h}{(2u+2h-1)(2u-1)}$

STEP 9

Now take the limit as $h$ approaches 0.
$g'(u) = \frac{-2u}{(2u-1)^2}$

STEP 10

The derivative of the function $g(u)$ is:
$g'(u) = \frac{-2u}{(2u-1)^2}$

STEP 11

To find the domain of the function $g(u)$ , we need to determine where the function is defined. The function is undefined where the denominator is zero.
$2u - 1 = 0$

STEP 12

Solve for $u$ to find the value that is not in the domain of $g(u)$ .
$u = \frac{1}{2}$

STEP 13

The domain of $g(u)$ is all real numbers except $u = \frac{1}{2}$ .
$\text{Domain of } g(u): (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

STEP 14

To find the domain of the derivative $g'(u)$ , we look at the denominator of the derivative.
$(2u-1)^2 = 0$

STEP 15

Solve for $u$ to find the value that is not in the domain of $g'(u)$ .
$u = \frac{1}{2}$

STEP 16

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the derivative of the function g(u)=u+12u−1g(u) = \frac{u+1}{2u-1}g(u)=2u−1u+1​ using the definition of derivative. State the domain of g(u)g(u)g(u) and its derivative in interval notation.

STEP 1

STEP 2

First, we need to find the expression for g(u+h) g(u+h) g(u+h) by substituting u+h u+h u+h into the function g(u) g(u) g(u).g(u+h)=(u+h)+12(u+h)−1 g(u+h) = \frac{(u+h)+1}{2(u+h)-1} g(u+h)=2(u+h)−1(u+h)+1​

STEP 3

Simplify the expression for g(u+h) g(u+h) g(u+h).g(u+h)=u+h+12u+2h−1 g(u+h) = \frac{u+h+1}{2u+2h-1} g(u+h)=2u+2h−1u+h+1​

STEP 4

Now, we will set up the difference quotient for the definition of the derivative.g′(u)=lim⁡h→0u+h+12u+2h−1−u+12u−1h g'(u) = \lim_{h \to 0} \frac{\frac{u+h+1}{2u+2h-1} - \frac{u+1}{2u-1}}{h} g′(u)=h→0lim​h2u+2h−1u+h+1​−2u−1u+1​​

STEP 5

STEP 6

STEP 7

Simplify the numerator by combining like terms.g′(u)=lim⁡h→0−2u−2h2h(2u+2h−1)(2u−1) g'(u) = \lim_{h \to 0} \frac{-2u - 2h^2}{h(2u+2h-1)(2u-1)} g′(u)=h→0lim​h(2u+2h−1)(2u−1)−2u−2h2​

STEP 8

Cancel out the h h h in the numerator with the h h h in the denominator.g′(u)=lim⁡h→0−2u−2h(2u+2h−1)(2u−1) g'(u) = \lim_{h \to 0} \frac{-2u - 2h}{(2u+2h-1)(2u-1)} g′(u)=h→0lim​(2u+2h−1)(2u−1)−2u−2h​

STEP 9

Now take the limit as h h h approaches 0.g′(u)=−2u(2u−1)2 g'(u) = \frac{-2u}{(2u-1)^2} g′(u)=(2u−1)2−2u​

STEP 10

The derivative of the function g(u) g(u) g(u) is:g′(u)=−2u(2u−1)2 g'(u) = \frac{-2u}{(2u-1)^2} g′(u)=(2u−1)2−2u​

STEP 11

To find the domain of the function g(u) g(u) g(u), we need to determine where the function is defined. The function is undefined where the denominator is zero.2u−1=0 2u - 1 = 0 2u−1=0

STEP 12

Solve for u u u to find the value that is not in the domain of g(u) g(u) g(u).u=12 u = \frac{1}{2} u=21​

STEP 13

The domain of g(u) g(u) g(u) is all real numbers except u=12 u = \frac{1}{2} u=21​.Domain of g(u):(−∞,12)∪(12,∞) \text{Domain of } g(u): (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty) Domain of g(u):(−∞,21​)∪(21​,∞)

STEP 14

To find the domain of the derivative g′(u) g'(u) g′(u), we look at the denominator of the derivative.(2u−1)2=0 (2u-1)^2 = 0 (2u−1)2=0

STEP 15

Solve for u u u to find the value that is not in the domain of g′(u) g'(u) g′(u).u=12 u = \frac{1}{2} u=21​

STEP 16

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the derivative of the function $g(u) = \frac{u+1}{2u-1}$ using the definition of derivative. State the domain of $g(u)$ and its derivative in interval notation.

First, we need to find the expression for $g(u+h)$ by substituting $u+h$ into the function $g(u)$ .
$g(u+h) = \frac{(u+h)+1}{2(u+h)-1}$

Simplify the expression for $g(u+h)$ .
$g(u+h) = \frac{u+h+1}{2u+2h-1}$

Now, we will set up the difference quotient for the definition of the derivative.
$g'(u) = \lim_{h \to 0} \frac{\frac{u+h+1}{2u+2h-1} - \frac{u+1}{2u-1}}{h}$

Simplify the numerator by combining like terms.
$g'(u) = \lim_{h \to 0} \frac{-2u - 2h^2}{h(2u+2h-1)(2u-1)}$

Cancel out the $h$ in the numerator with the $h$ in the denominator.
$g'(u) = \lim_{h \to 0} \frac{-2u - 2h}{(2u+2h-1)(2u-1)}$

Now take the limit as $h$ approaches 0.
$g'(u) = \frac{-2u}{(2u-1)^2}$

The derivative of the function $g(u)$ is:
$g'(u) = \frac{-2u}{(2u-1)^2}$

To find the domain of the function $g(u)$ , we need to determine where the function is defined. The function is undefined where the denominator is zero.
$2u - 1 = 0$

Solve for $u$ to find the value that is not in the domain of $g(u)$ .
$u = \frac{1}{2}$

The domain of $g(u)$ is all real numbers except $u = \frac{1}{2}$ .
$\text{Domain of } g(u): (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

To find the domain of the derivative $g'(u)$ , we look at the denominator of the derivative.
$(2u-1)^2 = 0$

Solve for $u$ to find the value that is not in the domain of $g'(u)$ .
$u = \frac{1}{2}$