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hichkok12 [17]
3 years ago
8

Find the differential of each function. (a) y = tan( 7t ) dy = Incorrect: Your answer is incorrect. (b) y = 4 − v2 4 + v2 dy =

Mathematics
1 answer:
Kisachek [45]3 years ago
6 0

Answer:

(a) dy= 7sec^2(7t)dt\\\\(b) dy = \frac{-16v}{(4+v^2)^2} dv

Step-by-step explanation:

Given;

(a) y = tan (7t)

let u = 7t

y = tan(u)

du/dt = 7

dy/du = sec²(u)

\frac{dy}{dt} = \frac{dy}{du} *\frac{du}{dt}\\\\\frac{dy}{dt} = sec^2(u)* 7\\\\\frac{dy}{dt} =7sec^2(u)\\\\\frac{dy}{dt} = 7sec^2(7t)\\\\dy = 7sec^2(7t)dt

(b)  

y = \frac{4-v^2}{4+v^2}\\\\

let u = 4 - v²

du/dv = -2v

let v = 4 + v²

dv/du = 2v

y = \frac{4-v^2}{4+v^2}\\\\\frac{dy}{dv} = \frac{vdu-udv}{v^2} \\\\\frac{dy}{dv} =\frac{-2v(4+v^2)-2v(4-v^2)}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-8v-2v^3-8v+2v^3}{(4+v^2)^2}\\\\\frac{dy}{dv} =\frac{-16v}{(4+v^2)^2}\\\\dy = \frac{-16v}{(4+v^2)^2}dv

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Step-by-step explanation:

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25 (a square) - 4 (b square) + 28bc - 49 (c square)
LUCKY_DIMON [66]

Answer:

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Step-by-step explanation:

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Thus, the original  25a^2 - 4b^2 + 28bc - 49c^2  looks like:

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Recall that a^2 - b^2 is a special product, the product of (a + b) and (a - b).  Applying this pattern to the problem at hand, we conclude:

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