Equation of a Tangent Line

Question

I am asked to give the equation of the tangent line to the curve

$g(x) = {3x\over \sqrt{x^2 +1}}$ at the point $(1, g(1))$

I understand to find the equation I need to solve for $g(1) = y$ and solve $g'(x)$ to find the slope of the line. Once I have these two numbers I can create my equation $y = m x +b$.

The problem I am having is the numbers I have come up with have me questioning my calculations. $$ g(1) = {3\sqrt{2} \over 2} \qquad \text{and} \qquad g'(1) = -{3(x) \over (x^2 + 1)^{3/2}} \approx 1.06. $$

Answer 1

The equation of the tangent to a curve at the point $(x_0,y_0)$, is given by: $$y = g'(x_0) \left( x-x_0 \right) + y_0$$ in your case $x_0=1$ and $y_0 = g(1)$

Answer 2

$$g(x)={3x\over \sqrt{x^2+1}}\implies g'(x)={\sqrt{x^2+1}\cdot3-{3x\cdot 2x\over 2\sqrt{x^2+1}}\over (\sqrt{x^2+1})^2}$$ Now put in the values and proceed..