Maharashtra State BoardHSC Commerce 12th Board Exam
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Find dydx, if y = x(x) + 20(x) Solution: Let y = x(x) + 20(x) Let u = x□ and v□ ∴ y = u + v Diff. w.r.to x, we get dydx=□dx+dv□ .....(i) Now, u = xx Taking log on both sides, we get log u = x × log x - Mathematics and Statistics

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Sum

Find `("d"y)/("d"x)`, if y = x(x) + 20(x) 

Solution: Let y = x(x) + 20(x) 

Let u = `x^square` and v = `square^x`

∴ y = u + v

Diff. w.r.to x, we get

`("d"y)/("d"x) = square/("d"x) + "dv"/square`   .....(i)

Now, u = xx

Taking log on both sides, we get

log u = x × log x

Diff. w.r.to x,

`1/"u"*"du"/("d"x) = x xx 1/square + log x xx square`

∴ `"du"/("d"x)` = u(1 + log x)

∴ `"du"/("d"x) = x^x (1 +  square)`    .....(ii)

Now, v = 20x

Diff.w.r.to x, we get

`"dv"/("d"x") = 20^square*log(20)`     .....(iii)

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

`("d"y)/("d"x)` = xx(1 + log x) + 20x.log(20)

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Solution

Let y = x(x) + 20(x) 

Let u = `x^x` and v = `20^x`

∴ y = u + v

Diff. w.r.to x, we get

`("d"y)/("d"x) = "du"/("d"x) + "dv"/("d"x)`   .....(i)

Now, u = xx

Taking log on both sides, we get

log u = x × log x

Diff. w.r.to x,

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

∴ `"du"/("d"x)` = u(1 + log x)

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

Now, v = 20x

Diff.w.r.to x, we get

`"dv"/("d"x") = 20^x*log(20)`     .....(iii)

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

`("d"y)/("d"x)` = xx(1 + log x) + 20x.log(20)

Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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