# Find dydx if y = xx+(7x-1)x - Mathematics and Statistics

Sum

Find "dy"/"dx" if y = "x"^"x" + ("7x" - 1)^"x"

#### Solution

y = "x"^"x" + ("7x" - 1)^"x"

Let u = xx  and v = ("7x" - 1)^"x"

∴ y = u + v

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

"dy"/"dx" = "du"/"dx" + "dv"/"dx" ....(i)

Now, u = xx

Taking logarithm of both sides, we get

log u = log(xx)

∴ log u = x. log x

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

1/"u" * "du"/"dx" = "x" * "d"/"dx" (log "x") + log "x" * "d"/"dx"(x)

= "x" * 1/"x" + log "x"  * (1)

∴ 1/"u" * "du"/"dx" = 1 + log x

∴ "du"/"dx" = u(1 + log x)

∴ "d"/"dx" ("x"^"x") = "x"^"x"(1 + log x)    ....(ii)

Also, v = (7x – 1)x

Taking logarithm of both sides, we get

log v = log(7x - 1)x

∴ log v = x. log(7x – 1)

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

1/"v" * "dv"/"dx" = "x" * "d"/"dx" log ("7x" - 1) + log ("7x" - 1) * "d"/"dx"(x)

= "x" * 1/("7x" - 1) * "d"/"dx" (7"x" - 1) + log (7"x" - 1) * (1)

∴ 1/"v" * "dv"/"dx" = "x"/(7"x" - 1) (7 - 0) + log (7"x" - 1)

∴ "dv"/"dx" = "v"["7x"/(7"x" - 1) + log(7"x" - 1)]

∴"dv"/"dx" = (7"x" - 1)^"x" ["7x"/(7"x" - 1) + log(7"x" - 1)]      ....(iii)

Substituting (ii) and (iii) in (i), we get

"dy"/"dx" = "x"^"x" (1 + log "x") + (7"x" - 1)^"x" [log(7"x" - 1) + "7x"/(7"x" - 1)]

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

#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 3 Differentiation
Miscellaneous Exercise 3 | Q 4.09 | Page 100
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