Question

Differentiate y = x5√x. SOLUTION Since both the base and the exponent are variable, we use logarithmic differentiation: ln y = ln x5√x = 5 x ln(x) y' y = 5 x · + (ln(x)) · y' = y + 5 ln(x) = x5√x . Another method is to write x5√x = eln(x) 5√x : d dx x5√x = d dx = d dx 5 x ln(x) = x5√x .

Answer 1

`y'=x^(5√(x) ) ((5)/(√(x) )+(5)/(2√(x) )lnx)`

Explanation:

Given:
`y=x^(5√(x) )`

Take ln of both sides

`lny=lnx^(5√(x) ) = 5√(x)lnx\\\\lny=5√(x)lnx`

Differentiate with respect to x

`(d)/(dx)(lny)=(d)/(dx)(5√(x)lnx)\\\\(d)/(dy)(lny)*(dy)/(dx)=5√(x)(d)/(dx)lnx+(d)/(dx)5√(x)*lnx)\\\\(1)/(y)*(dy)/(dx)=(5√(x)*(1)/(x))+(5*(1)/(2)x^{-(1)/(2)}lnx)\\ \\(1)/(y)*y^(')=(5)/(√(x) )+(5)/(2√(x) )lnx \\\\y'=y((5)/(√(x) )+(5)/(2√(x) )lnx)\\\\y'=x^(5√(x) ) ((5)/(√(x) )+(5)/(2√(x) )lnx)`