Solved on Feb 10, 2024

Find the derivative of $y = \frac{4t - 3}{8t + 1}$ using the quotient rule. The solution is $\frac{d y}{d t} = \frac{32}{(8t + 1)^2}$ .

STEP 1

Assumptions
1. The function given is $y = \frac{4t - 3}{8t + 1}$ .
2. We will use the quotient rule to find the derivative of the function with respect to $t$ .
3. The quotient rule states that if $y = \frac{u(t)}{v(t)}$ , then the derivative $\frac{dy}{dt} = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$ , where $u(t)$ and $v(t)$ are differentiable functions of $t$ .

STEP 2

Identify the numerator $u(t)$ and the denominator $v(t)$ of the function.
$u(t) = 4t - 3$ $v(t) = 8t + 1$

STEP 3

Find the derivative of the numerator $u(t)$ with respect to $t$ , denoted as $u'(t)$ .
$u'(t) = \frac{d}{dt}(4t - 3)$

STEP 4

Calculate $u'(t)$ .
$u'(t) = 4$

STEP 5

Find the derivative of the denominator $v(t)$ with respect to $t$ , denoted as $v'(t)$ .
$v'(t) = \frac{d}{dt}(8t + 1)$

STEP 6

Calculate $v'(t)$ .
$v'(t) = 8$

STEP 7

Apply the quotient rule to find $\frac{dy}{dt}$ .
$\frac{dy}{dt} = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$

STEP 8

Substitute $u(t)$ , $v(t)$ , $u'(t)$ , and $v'(t)$ into the quotient rule formula.
$\frac{dy}{dt} = \frac{4(8t + 1) - (4t - 3)8}{(8t + 1)^2}$

STEP 9

Distribute $u'(t)$ into $v(t)$ and $u(t)$ into $v'(t)$ .
$\frac{dy}{dt} = \frac{32t + 4 - 32t + 24}{(8t + 1)^2}$

STEP 10

Combine like terms in the numerator.
$\frac{dy}{dt} = \frac{4 + 24}{(8t + 1)^2}$

STEP 11

Add the constants in the numerator.
$\frac{dy}{dt} = \frac{28}{(8t + 1)^2}$
The derivative of the function $y = \frac{4t - 3}{8t + 1}$ with respect to $t$ is:
$\frac{dy}{dt} = \frac{28}{(8t + 1)^2}$

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Find the derivative of $y = \frac{4t - 3}{8t + 1}$ using the quotient rule. The solution is $\frac{d y}{d t} = \frac{32}{(8t + 1)^2}$ .

STEP 1

STEP 2

Identify the numerator $u(t)$ and the denominator $v(t)$ of the function.
$u(t) = 4t - 3$ $v(t) = 8t + 1$

STEP 3

Find the derivative of the numerator $u(t)$ with respect to $t$ , denoted as $u'(t)$ .
$u'(t) = \frac{d}{dt}(4t - 3)$

STEP 4

Calculate $u'(t)$ .
$u'(t) = 4$

STEP 5

Find the derivative of the denominator $v(t)$ with respect to $t$ , denoted as $v'(t)$ .
$v'(t) = \frac{d}{dt}(8t + 1)$

STEP 6

Calculate $v'(t)$ .
$v'(t) = 8$

STEP 7

Apply the quotient rule to find $\frac{dy}{dt}$ .
$\frac{dy}{dt} = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$

STEP 8

Substitute $u(t)$ , $v(t)$ , $u'(t)$ , and $v'(t)$ into the quotient rule formula.
$\frac{dy}{dt} = \frac{4(8t + 1) - (4t - 3)8}{(8t + 1)^2}$

STEP 9

Distribute $u'(t)$ into $v(t)$ and $u(t)$ into $v'(t)$ .
$\frac{dy}{dt} = \frac{32t + 4 - 32t + 24}{(8t + 1)^2}$

STEP 10

Combine like terms in the numerator.
$\frac{dy}{dt} = \frac{4 + 24}{(8t + 1)^2}$

STEP 11

Add the constants in the numerator.
$\frac{dy}{dt} = \frac{28}{(8t + 1)^2}$
The derivative of the function $y = \frac{4t - 3}{8t + 1}$ with respect to $t$ is:
$\frac{dy}{dt} = \frac{28}{(8t + 1)^2}$

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Find the derivative of y=4t−38t+1y = \frac{4t - 3}{8t + 1}y=8t+14t−3​ using the quotient rule. The solution is dydt=32(8t+1)2\frac{d y}{d t} = \frac{32}{(8t + 1)^2}dtdy​=(8t+1)232​.

STEP 1

STEP 2

Identify the numerator u(t) u(t) u(t) and the denominator v(t) v(t) v(t) of the function.u(t)=4t−3 u(t) = 4t - 3 u(t)=4t−3 v(t)=8t+1 v(t) = 8t + 1 v(t)=8t+1

STEP 3

Find the derivative of the numerator u(t) u(t) u(t) with respect to t t t, denoted as u′(t) u'(t) u′(t).u′(t)=ddt(4t−3) u'(t) = \frac{d}{dt}(4t - 3) u′(t)=dtd​(4t−3)

STEP 4

Calculate u′(t) u'(t) u′(t).u′(t)=4 u'(t) = 4 u′(t)=4

STEP 5

Find the derivative of the denominator v(t) v(t) v(t) with respect to t t t, denoted as v′(t) v'(t) v′(t).v′(t)=ddt(8t+1) v'(t) = \frac{d}{dt}(8t + 1) v′(t)=dtd​(8t+1)

STEP 6

Calculate v′(t) v'(t) v′(t).v′(t)=8 v'(t) = 8 v′(t)=8

STEP 7

Apply the quotient rule to find dydt \frac{dy}{dt} dtdy​.dydt=u′(t)v(t)−u(t)v′(t)[v(t)]2 \frac{dy}{dt} = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} dtdy​=[v(t)]2u′(t)v(t)−u(t)v′(t)​

STEP 8

Substitute u(t) u(t) u(t), v(t) v(t) v(t), u′(t) u'(t) u′(t), and v′(t) v'(t) v′(t) into the quotient rule formula.dydt=4(8t+1)−(4t−3)8(8t+1)2 \frac{dy}{dt} = \frac{4(8t + 1) - (4t - 3)8}{(8t + 1)^2} dtdy​=(8t+1)24(8t+1)−(4t−3)8​

STEP 9

Distribute u′(t) u'(t) u′(t) into v(t) v(t) v(t) and u(t) u(t) u(t) into v′(t) v'(t) v′(t).dydt=32t+4−32t+24(8t+1)2 \frac{dy}{dt} = \frac{32t + 4 - 32t + 24}{(8t + 1)^2} dtdy​=(8t+1)232t+4−32t+24​

STEP 10

Combine like terms in the numerator.dydt=4+24(8t+1)2 \frac{dy}{dt} = \frac{4 + 24}{(8t + 1)^2} dtdy​=(8t+1)24+24​

STEP 11

Add the constants in the numerator.dydt=28(8t+1)2 \frac{dy}{dt} = \frac{28}{(8t + 1)^2} dtdy​=(8t+1)228​The derivative of the function y=4t−38t+1 y = \frac{4t - 3}{8t + 1} y=8t+14t−3​ with respect to t t t is:dydt=28(8t+1)2 \frac{dy}{dt} = \frac{28}{(8t + 1)^2} dtdy​=(8t+1)228​

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Find the derivative of $y = \frac{4t - 3}{8t + 1}$ using the quotient rule. The solution is $\frac{d y}{d t} = \frac{32}{(8t + 1)^2}$ .

Identify the numerator $u(t)$ and the denominator $v(t)$ of the function.
$u(t) = 4t - 3$ $v(t) = 8t + 1$

Find the derivative of the numerator $u(t)$ with respect to $t$ , denoted as $u'(t)$ .
$u'(t) = \frac{d}{dt}(4t - 3)$

Calculate $u'(t)$ .
$u'(t) = 4$

Find the derivative of the denominator $v(t)$ with respect to $t$ , denoted as $v'(t)$ .
$v'(t) = \frac{d}{dt}(8t + 1)$

Calculate $v'(t)$ .
$v'(t) = 8$

Apply the quotient rule to find $\frac{dy}{dt}$ .
$\frac{dy}{dt} = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$

Substitute $u(t)$ , $v(t)$ , $u'(t)$ , and $v'(t)$ into the quotient rule formula.
$\frac{dy}{dt} = \frac{4(8t + 1) - (4t - 3)8}{(8t + 1)^2}$

Distribute $u'(t)$ into $v(t)$ and $u(t)$ into $v'(t)$ .
$\frac{dy}{dt} = \frac{32t + 4 - 32t + 24}{(8t + 1)^2}$

Combine like terms in the numerator.
$\frac{dy}{dt} = \frac{4 + 24}{(8t + 1)^2}$

Add the constants in the numerator.
$\frac{dy}{dt} = \frac{28}{(8t + 1)^2}$
The derivative of the function $y = \frac{4t - 3}{8t + 1}$ with respect to $t$ is:
$\frac{dy}{dt} = \frac{28}{(8t + 1)^2}$