Find dydx if y = (3x-4)3(x + 1)4(x + 2) - Mathematics and Statistics

Sum

Find "dy"/"dx" if y = sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2")))

Solution

y = sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2")))

= ("3x" - 4)^(3/2)/(("x + 1")^(4/2)*("x + 2")^(1/2))

Taking logarithm of both sides, we get

log y = log[("3x" - 4)^(3/2)/(("x + 1")^(4/2)*("x + 2")^(1/2))]

= log ("3x" - 4)^(3/2) - [log ("x + 1")^2 + log ("x + 2")^(1/2)]

= 3/2 log("3x" - 4) - 2log ("x + 1") - 1/2 log ("x + 2")

Differentiating both sides w.r.t. x, we get

1/"y" * "dy"/"dx" = 3/2 * "d"/"dx" [log (3"x" - 4)] - 2 "d"/"dx" [log ("x + 1")] - 1/2 * "d"/"dx" [log ("x + 2")]

= 3/2 * 1/("3x" - 4) * "d"/"dx" ("3x" - 4) - 2 * 1/("x + 1") * "d"/"dx" ("x + 1") - 1/2 * 1/("x + 2") * "d"/"dx" ("x + 2")

∴ 1/"y" * "dy"/"dx" = 3/(2("3x - 4")) xx 3 - 2/("x + 1") xx 1 - 1/(2 ("x + 2")) xx 1

∴ 1/"y" * "dy"/"dx" = 9/(2("3x - 4")) - 2/("x + 1") - 1/(2 ("x + 2"))

∴ "dy"/"dx" = "y"/2 [9/"3x - 4" - 4/"x + 1" - 1/"x + 2"]

∴ "dy"/"dx" = 1/2 sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2"))) [9/"3x - 4" - 4/"x + 1" - 1/"x + 2"]

Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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Chapter 3: Differentiation - Miscellaneous Exercise 3 [Page 100]

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Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 3 Differentiation
Miscellaneous Exercise 3 | Q 4.08 | Page 100
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