Maharashtra State BoardHSC Commerce 12th Board Exam
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Differentiate e4x+5 with respect to 104x. - Mathematics and Statistics

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Sum

Differentiate `"e"^("4x" + 5)` with respect to 104x.

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Solution

Let u = `"e"^(("4x" + 5))` and v = 104x.

u = `"e"^(("4x" + 5))`

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

`"du"/"dx" = "e"^(("4x" + 5)) * "d"/"dx" (4"x" + 5)`

`= "e"^(("4x" + 5)) * (4 + 0)`

∴ `"du"/"dx" = 4 * "e"^(("4x" + 5)) *`

v = 104x 

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

`"dv"/"dx" = 10^"4x" * log 10 * "d"/"dx" ("4x")`

∴ `"dv"/"dx" = 10^"4x" * (log 10) (4)`

∴ `"du"/"dv" = ("du"/"dx")/("dv"/"dx") = (4 * "e"^(("4x" + 5)))/(10^"4x" * (log 10)(4))`

∴ `"du"/"dv" = ("e"^(("4x" + 5)))/(10^"4x" * (log 10)`

Concept: Derivatives of Composite Functions - Chain Rule
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 3 Differentiation
Miscellaneous Exercise 3 | Q 4.18 | Page 101
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