# How do you differentiate y= sqrt(x) e^(x^2) (x^2+3)^5?

Sep 10, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sqrt{x} {e}^{{x}^{2}} {\left({x}^{2} + 3\right)}^{5}\right) \left(\frac{1}{2 x} + 2 x + \frac{10 x}{{x}^{2} + 3}\right)$

#### Explanation:

I would use logarithmic differentiation.

lny = ln(sqrt(x)e^(x^2)(x^2 + 3)^5))

Using $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$.

$\ln y = \ln \sqrt{x} + \ln \left({e}^{{x}^{2}}\right) + \ln {\left({x}^{2} + 3\right)}^{5}$

$\ln y = \ln {x}^{\frac{1}{2}} + \ln \left({e}^{{x}^{2}}\right) + \ln {\left({x}^{2} + 3\right)}^{5}$

Now we use $\ln {a}^{n} = n \ln a$.

$\ln y = \frac{1}{2} \ln x + {x}^{2} \ln \left(e\right) + 5 \ln \left({x}^{2} + 3\right)$

$\ln y = \frac{1}{2} \ln x + {x}^{2} + 5 \ln \left({x}^{2} + 3\right)$

Now the derivative is given by the chain rules and $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$.

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{1}{2 x} + 2 x + \frac{5 \left(2 x\right)}{{x}^{2} + 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{1}{2 x} + 2 x + \frac{10 x}{{x}^{2} + 3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sqrt{x} {e}^{{x}^{2}} {\left({x}^{2} + 3\right)}^{5}\right) \left(\frac{1}{2 x} + 2 x + \frac{10 x}{{x}^{2} + 3}\right)$

Hopefully this helps!