# Question c0a68

Nov 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x} \log \left(\sqrt{x}\right) \ln 10 + \sqrt{x}}{2 x \ln 10}$

#### Explanation:

$y = \sqrt{x} \log \sqrt{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}} \log \sqrt{x} + \frac{\sqrt{x}}{\sqrt{x} \ln 10} \cdot \frac{1}{2 \sqrt{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}} \log \sqrt{x} + \frac{\cancel{\sqrt{x}}}{\sqrt{x} \ln 10} \cdot \frac{1}{2 \cancel{\sqrt{x}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \log \frac{\sqrt{x}}{2 \sqrt{x}} + \frac{1}{2 \sqrt{x} \ln 10}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x} \log \left(\sqrt{x}\right) \ln 10 + \sqrt{x}}{2 x \ln 10}$

Nov 4, 2016

$\frac{\sqrt{x} + \ln 10 \cdot \sqrt{x} \cdot \log \sqrt{x}}{2 \ln 10 \cdot x}$

#### Explanation:

Given the expression
$y = \sqrt{x} \log \sqrt{x}$
Rewriting it as exponents
$y = {x}^{\frac{1}{2}} \log {x}^{\frac{1}{2}}$
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
$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$ and $\frac{d}{\mathrm{dx}} {\log}_{10} x = \frac{1}{x \ln 10}$.
Therefore we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\frac{1}{2}} \times \left(\frac{1}{{x}^{\frac{1}{2}} \ln 10} \cdot \frac{1}{2} {x}^{\frac{1}{2} - 1}\right) + \log {x}^{\frac{1}{2}} \times \frac{1}{2} {x}^{\frac{1}{2} - 1}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cancel{{x}^{\frac{1}{2}}} \left(\frac{1}{\cancel{{x}^{\frac{1}{2}}} \ln 10} \cdot \frac{1}{2} {x}^{- \frac{1}{2}}\right) + \log {x}^{\frac{1}{2}} \times \frac{1}{2} {x}^{- \frac{1}{2}}$
Reverting back to sqrt# representation we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln 10} \cdot \frac{1}{2} \frac{1}{\sqrt{x}} + \frac{1}{2} \frac{1}{\sqrt{x}} \log \sqrt{x}$
Simplifying we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \ln 10 \cdot \sqrt{x}} + \log \frac{\sqrt{x}}{2 \sqrt{x}}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + \ln 10 \cdot \log \sqrt{x}}{2 \ln 10 \cdot \sqrt{x}}$
Rationalizing the denominator

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{x} + \ln 10 \cdot \sqrt{x} \cdot \log \sqrt{x}}{2 \ln 10 \cdot x}$