# 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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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)

#### 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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Chapter 1.3: Differentiation - Q.6
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