Question

Question #c0a68

Answer 1

`dy/dx=(sqrtx log(sqrtx)ln10+sqrtx)/(2x ln10)`

Explanation:

`y=sqrtx log sqrtx`

`dy/dx=1/(2sqrtx) log sqrtx+sqrtx/(sqrtxln10)*1/(2sqrtx)`

`dy/dx=1/(2sqrtx) log sqrtx+cancel sqrtx/(sqrtxln10)*1/(2cancelsqrtx)`

`dy/dx= log sqrtx/(2sqrtx)+1/(2sqrtxln10)`

`dy/dx=(sqrtx log(sqrtx)ln10+sqrtx)/(2x ln10)`

Answer 2

`(sqrtx+ln10*sqrtx* logsqrtx)/(2 ln10*x)`

Explanation:

Given the expression
`y=sqrtx log sqrtx`
Rewriting it as exponents
`y=x^(1/2) log x^(1/2)`
We need to use Product rule, which can be stated as:

"The derivative of a product of two functions is the sum of first function times the derivative of the second, and second function times the derivative of the first."
As the base of `log` function has not been explicitly given, assuming it to equal to `10`. We also know that
`d/dx x^n=nx^(n-1)` and `d/dxlog_10 x=1/(xln10)`.
Therefore we get

`dy/dx=x^(1/2) xx(1/(x^(1/2)ln10)*1/2 x^(1/2-1))+log x^(1/2)xx1/2 x^(1/2-1)`
`=>dy/dx=cancel(x^(1/2))(1/(cancel(x^(1/2))ln10)*1/2 x^(-1/2))+log x^(1/2)xx1/2 x^(-1/2)`
Reverting back to `sqrt` representation we get

`dy/dx=1/(ln10)*1/2 1/sqrtx+1/2 1/sqrtxlog sqrtx`
Simplifying we get

`dy/dx= 1/(2ln10*sqrtx)+log sqrtx/(2sqrtx)`
`=>dy/dx= (1+ln10*log sqrtx)/(2ln10*sqrtx)`
Rationalizing the denominator

`dy/dx=(sqrtx+ln10*sqrtx* logsqrtx)/(2 ln10*x)`