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