Solve 1/sqrt{3+2x}

$$\frac { 1 } { \sqrt { 3 } + 2 x }$$

Evaluate

$$\frac{1}{2x+\sqrt{3}}$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}+2x} by multiplying numerator and denominator by \sqrt{3}-2x.

$$\frac{\sqrt{3}-2x}{\left(\sqrt{3}+2x\right)\left(\sqrt{3}-2x\right)}$$

Consider \left(\sqrt{3}+2x\right)\left(\sqrt{3}-2x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{3}-2x}{\left(\sqrt{3}\right)^{2}-\left(2x\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{3}-2x}{3-\left(2x\right)^{2}}$$

Expand \left(2x\right)^{2}.

$$\frac{\sqrt{3}-2x}{3-2^{2}x^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\sqrt{3}-2x}{3-4x^{2}}$$

Differentiate w.r.t. x

$$-\frac{2}{\left(2x+\sqrt{3}\right)^{2}}$$