Question

Find `("d"y)/("d"x)`, if y = x^x + (7x – 1)^x 

Answer 1

y = x^x + (7x – 1)^x 

Let u = x^x and v = (7x – 1)^x 

∴ y = u + v

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

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

Now, u = x^x

Taking logarithm of both sides, we get

log u = log (x^x)

∴ log u = x. log x

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

`"d"/("d"x)(log "u") = x*"d"/("d"x)(log x) + logx*"d"/("d"x)(x)`

∴ `1/"u"*"du"/("d"x) = x*1/x + logx*1`

∴ `1/"u"*"du"/("d"x)` = 1 + log x

∴ `"du"/("d"x)` = u(1 + log x)

∴ `"du"/("d"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

`"d"/("d"x)(log "v") = x*"d"/("d"x)[log(7x - 1)] + log(7x - 1)*"d"/("d"x)(x)`

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

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

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

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

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

`("d"y)/("d"x) = x^x(1 + logx) + (7x - 1)^x[log(7x - 1) + (7x)/(7x - 1)]`