# How do you differentiate y= ((x+3)(x^2+1)^3(x+1)^2)/(x^2+10)^(1/2)?

##### 1 Answer
Jan 10, 2018

dy/dx= ((x + 3)(x^2 + 1)^3(x + 1)^2)/(x^2 + 10)^(1/2))(1/(x+ 3) + (6x)/(x^2 +1) + 2/(x + 1) - x/(x^2 + 10))

#### Explanation:

Use logarithmic differentiation.

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

Use the logarithm laws that state $\ln \left(a b\right) = \ln a + \ln b$ and $\ln \left(\frac{a}{b}\right) = \ln a - \ln b$.

$\ln y = \ln \left(x + 3\right) + \ln {\left({x}^{2} + 1\right)}^{3} + \ln {\left(x + 1\right)}^{2} - \ln {\left({x}^{2} + 10\right)}^{\frac{1}{2}}$

Now use $\ln \left({a}^{n}\right) = n \ln a$.

$\ln y = \ln \left(x + 3\right) + 3 \ln \left({x}^{2} + 1\right) + 2 \ln \left(x + 1\right) - \frac{1}{2} \ln \left({x}^{2} + 10\right)$

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

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{1}{x + 3} + \frac{6 x}{{x}^{2} + 1} + \frac{2}{x + 1} - \frac{x}{{x}^{2} + 10}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\frac{1}{x + 3} + \frac{6 x}{{x}^{2} + 1} + \frac{2}{x + 1} - \frac{x}{{x}^{2} + 10}\right)$

dy/dx= ((x + 3)(x^2 + 1)^3(x + 1)^2)/(x^2 + 10)^(1/2))(1/(x+ 3) + (6x)/(x^2 +1) + 2/(x + 1) - x/(x^2 + 10))

Hopefully this helps!