Question

Use logarithmic differentiation to find the derivative of the function. y = sqrt(x)e^x^2(x^2 + 5)^12

Answer 1

$y=\sqrt x\,e^{x^2}(x^2+5)^{12}=x^{1/2}e^{x^2}(x^2+5)^{1/2}$

Take the logarithm of both sides and expand the right hand side:

$\ln y=\ln\left(x^{1/2}e^{x^2}(x^2+5)^{1/2}\right)$

$\ln y=\ln x^{1/2}+\ln e^{x^2}+\ln(x^2+5)^{12}$

$\ln y=\dfrac12\ln x+x^2\ln e+12\ln(x^2+5)$

$\ln y=\dfrac12\ln x+x^2+12\ln(x^2+5)$

Now take the derivative of both sides with respect to $x$:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2x}+2x+\dfrac{24x}{x^2+5}$

$\dfrac{\mathrm dy}{\mathrm dx}=\left(\dfrac1{2x}+2x+\dfrac{24x}{x^2+5}\right)y$

$\dfrac{\mathrm dy}{\mathrm dx}=\left(\dfrac1{2x}+2x+\dfrac{24x}{x^2+5}\right)\sqrt x\,e^{x^2}(x^2+5)^{12}$

I'd stop there, but you could condense the right side a bit to get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4x^4+21x^2+48x+5}{2x(x^2+5)}\sqrt x\,e^{x^2}(x^2+5)^{12}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4x^4+21x^2+48x+5}{2\sqrt x}e^{x^2}(x^2+5)^{11}$