# How do you differentiate s(x)=log_3(x^2+5x)?

Jun 14, 2016

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

#### Explanation:

We can rewrite $s \left(x\right)$ using the change of base formula, which states that

${\log}_{a} \left(b\right) = {\log}_{c} \frac{b}{\log} _ c \left(a\right)$

Here, we will choose a base of $e$ because differentiation with the natural logarithm is simple.

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

When differentiating this, note that $s \left(x\right)$ is really the function $\ln \left({x}^{2} + 5 x\right)$ multiplied by the constant $\frac{1}{\ln} \left(3\right)$.

To find the derivative of $\ln \left({x}^{2} + 5 x\right)$, we will use the chain rule.

Since $\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$, we see that $\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot {u}^{'} .$

Thus:

${s}^{'} \left(x\right) = \frac{1}{\ln} \left(3\right) \cdot \frac{1}{{x}^{2} + 5 x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 5 x\right)$

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