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

Answer:

$y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$

Step-by-step explanation:

Given: $y=x^{5\sqrt{x} }$

Take ln of both sides

$lny=lnx^{5\sqrt{x} } = 5\sqrt{x}lnx\\\\lny=5\sqrt{x}lnx$

Differentiate with respect to x

$\frac{d}{dx}(lny)=\frac{d}{dx}(5\sqrt{x}lnx)\\\\\frac{d}{dy}(lny)*\frac{dy}{dx}=5\sqrt{x}\frac{d}{dx}lnx+\frac{d}{dx}5\sqrt{x}*lnx)\\\\\frac{1}{y}*\frac{dy}{dx}=(5\sqrt{x}*\frac{1}{x})+(5*\frac{1}{2}x^{-\frac{1}{2}}lnx)\\ \\\frac{1}{y}*y^{'}=\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx \\\\y'=y(\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)\\\\y'=x^{5\sqrt{x} } (\frac{5}{\sqrt{x} }+\frac{5}{2\sqrt{x} }lnx)$