Find an equation to the tangent line to a curve and parrallel

foreverlost

1. The problem statement, all variables and given/known data

Find an equation of the tangent line to the curve y = x√x that is parallel to the line
y = 1+3x.

2. Relevant equations

m = 3

3. The attempt at a solution

Here is my attempt: dy/dx(x) * dy/dx(x^(1/2)) = (1) * (1/2x^(-1/2)) = (1/2x^(-1/2))

(1/2x^(-1/2)) = 3 → x^(-1/2) = 6 → -√x = 6

x = -√36 = 6 → -6=6 -1=1

Thank you, I will not be able to respond for a bit as im leaving for home now.

LCKurtz

Two comments. First, why on earth would you not simplify ##x\sqrt x## to ##x^\frac 3 2## before you differentiate? And given that you didn't, the product rule for differentiation is not ##(fg)' = f'g'##, which is what you did. Try again.

Mark44

To add to what LCKurtz said about your oversimplification of the product rule, what you wrote does not mean what you think.
You don't take "dy/dx of x" or "dy/dx of x^(1/2)." You can take the derivative with respect to x of something, which you write as d/dx(x) + d/dx(x^(1/2)).

The expression dy/dx(x) means dy/dx times x, which I'm pretty sure isn't what you intended.