Find the slope of the tangent line for g(x)=x/(square root(5x+1); for x=1

Answer 1

For g(x) = x/√(5x+1) or x/(5x+1)^0.5, write g'(x) = {(5x+1)^0.5(dx/dx) − xd[(5x+1)^0.5]/dx} ÷ [(5x+1)^0.5]² which gives

{(5x+1)^0.5 − x(0.5×5)(5x+1)^(0.5-1)} ÷ (5x+1) or [(5x+1)^0.5/(5x+1)^0.5]{(5x+1)^0.5 − x(0.5×5)(5x+1)^(0.5-1)} ÷ (5x+1)

which reduces to {5x+1 − 2.5x}/(5x+1)^1.5 or {2.5x + 1}/(5x+1)^1.5.

{2.5x + 1}/(5x+1)^1.5, the derivative or slope of the line tangent to the graph of g(x) = x/√(5x+1) or x/(5x+1)^0.5,

is equal to (2.5+1)/6^1.5 or 3.5/(6√6) when x=1.

Using x equal to 1 in g(x) = x/√(5x+1) gives g(x) or y as 1/√6. Then place (x,y) equal to (1,1/√6) into y = [3.5/(6√6)]x + b to obtain 1/√6 = [3.5/(6√6)](1) + b or b = 2.5/(6√6). The equation of the line tangent to g(x) = x/√(5x+1) at (x,y) = (1,1/√6) is then obtained as y = [3.5/(6√6)]x + [2.5/(6√6)].