calculus problem?

Question

how do you dind the equation of the tangent line f(x)=x√(1+x^2) at (1,√2).

Answer 1

f (x) = (x) (1 + x²)^(1/2)
f `(x) = (1 + x²)^(1/2) + (1/2)(1 + x²)^(-1/2) (2x²)
f `(1) = 2^(1/2) + (1/2)(2)^(-1/2) (2)
f `(1) = 2^(1/2) + 2^(-1/2)
f `(1) = 2^(-1/2) (2 + 1)
f `(1) = 3 / √2

y - √2 = ( 3 / √2 )( x - 1 )
y = (3 /√2 ) x + (√2 - 3 / √2)
y = (3 / √2 ) x + (2 - 3√2) / 2
y = [ (3√2) / 2 ] x + (2 - 3√2) / 2

Answer 2

You would find d f(x)/dx as a function and evaluate it at x=1. The tangent would have a slope equal to
(- 1/ f'(1) ) and a y-intercept that would allow x to =1 when y=sqrt(2).

Answer 3

f '(1) is the slope of the line tangent to the graph of y = f(x) at the point (1, f(1)) = (1, √2).

Using point-slope form of a line ( y = m(x - a) + b ), the equation of the tangent line is

y = f '(1)(x - 1) + √2

To find f '(x) you can use the product and chain rules.