Solved on Feb 10, 2024

Find the derivative of y=4t38t+1y = \frac{4t - 3}{8t + 1} using the quotient rule. The solution is dydt=32(8t+1)2\frac{d y}{d t} = \frac{32}{(8t + 1)^2}.

STEP 1

Assumptions
1. The function given is y=4t38t+1 y = \frac{4t - 3}{8t + 1} .
2. We will use the quotient rule to find the derivative of the function with respect to t t .
3. The quotient rule states that if y=u(t)v(t) y = \frac{u(t)}{v(t)} , then the derivative 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} , where u(t) u(t) and v(t) v(t) are differentiable functions of t t .

STEP 2

Identify the numerator u(t) u(t) and the denominator v(t) v(t) of the function.
u(t)=4t3 u(t) = 4t - 3 v(t)=8t+1 v(t) = 8t + 1

STEP 3

Find the derivative of the numerator u(t) u(t) with respect to t t , denoted as u(t) u'(t) .
u(t)=ddt(4t3) u'(t) = \frac{d}{dt}(4t - 3)

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

Apply the quotient rule to find dydt \frac{dy}{dt} .
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}

STEP 8

Substitute u(t) u(t) , v(t) v(t) , u(t) u'(t) , and v(t) v'(t) into the quotient rule formula.
dydt=4(8t+1)(4t3)8(8t+1)2 \frac{dy}{dt} = \frac{4(8t + 1) - (4t - 3)8}{(8t + 1)^2}

STEP 9

Distribute u(t) u'(t) into v(t) v(t) and u(t) u(t) into v(t) v'(t) .
dydt=32t+432t+24(8t+1)2 \frac{dy}{dt} = \frac{32t + 4 - 32t + 24}{(8t + 1)^2}

STEP 10

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

STEP 11

Add the constants in the numerator.
dydt=28(8t+1)2 \frac{dy}{dt} = \frac{28}{(8t + 1)^2}
The derivative of the function y=4t38t+1 y = \frac{4t - 3}{8t + 1} with respect to t t is:
dydt=28(8t+1)2 \frac{dy}{dt} = \frac{28}{(8t + 1)^2}

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