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

[foreverlost]

Homework Statement

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

Homework Equations

m = 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 I am leaving for home now.

 

[LCKurtz]

Homework Statement

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

Homework Equations

m = 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 I am leaving for home now.

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.

dy/dx(x) * dy/dx(x^(1/2))

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.