Question

Hi there. This is my question: Find an equation of the tangent line to the curve y = x * [x^(1/2)] that is parallel to the the line y = 1 + 3x. I'm able to find the derivative which I have as 1.5x^(1/2) but don't know where to go from there.

Answer 1

\[y = x \sqrt{x}\]

Answer 2

\[y' = (3/2) \sqrt{x}\]

Answer 3

Yes, first you would need to find the DERIVATIVE of the curve. However you need to use the PRODUCT rule to find the derivative of \[x\]

Answer 4

You need to use the PRODUCT rule to find the derivative of \[x \sqrt{x}\]

Answer 5

do you know the product rule?

Answer 6

you don't need to product rule, by using index laws it can be simplified

\[y = x \sqrt{x} = x^1 \times x^{\frac{1}{2}} = x^{\frac{3}{2}}\]

so you have

\[y = x^{\frac{3}{2}}\]

so this makes finding the derivative so much easier

Answer 7

yes, i get \[y' = (3/2)\sqrt{x}\]

Answer 8

@adziz is a smartass dont listen to him

Answer 9

ok... so your derivative is now equal to 3, which is the slope of the parallel line

so

\[3 = \frac{3}{2} \sqrt{x}\]

find x...

Answer 10

ok, right!

Answer 11

@adziz is a meanybutt

Answer 12

now when you have x, substitute it into the original equation to the find y and you have a point and a slope of 3 so you can find the equation of the tangent...

Answer 13

@adziz

Answer 14

ddd

Answer 15

sweet

Answer 16

dsfsdffsfdhhfdhf

Answer 17

got it. thanks everyone.