Question

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 = x^x

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 = 20^x

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)` = x^x(1 + log x) + 20^x.log(20)

Answer 1

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 = x^x

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 = 20^x

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)` = x^x(1 + log x) + 20^x.log(20)