Question

can someone help me with finding the equation of the tangent line to the graph of the function at the indicated point?
y=x^3*squareroot(x^3+3) at x=1

Answer 1

I know I can rewrite it first as y=x^3(x^3+3)^1/2

Answer 2

take the derivative
replace \(x\) by 1, i.e. find \(f'(1)\) that is your slope
then use the point slope formula

Answer 3

if i take the derivative would i get y=(3x^2)*1/2(X^3+3)^-1/2

Answer 4

you have
\[x^3\sqrt{x^3+3}\] derivative requires product and chain rule
get
\[f'(x)=3x^2\sqrt{x^2+3}+x^3\frac{3x^2}{2\sqrt{x^3+3}}\]

Answer 5

clean up with some algebra if you need, or just replace \(x\) by \(1\) to get the slope

Answer 6

oh okay i tried doing it with just the product rule before but i never applied the chain rule okay

Answer 7

the slope would be ... let me work it out sorry i am little slow

Answer 8

slope is 21/4?

Answer 9

\[f'(x)=3x^2\sqrt{x^2+3}+x^3\frac{3x^2}{2\sqrt{x^3+3}}\]
\[f'(1)=3\sqrt{4}+\frac{3}{2\sqrt{4}}\]
\[f'(1)=6+\frac{3}{4}\]

Answer 10

im working it out again i think i missed a number in missed a number

Answer 11

yeah i added a one somewhere before

Answer 12

calculus is easy, its the algebra that tends to throw ya ;)

Answer 13

yeah when i get stuck and then ask for help it usually doing the algebra to fast that gets me

Answer 14

so to find the equation of the line can i plug in the x value of 1 and slope of 27/4 and solve for b? oh but i need to plug in x to the original equation first right? to get he y value?

Answer 15

there are many "forms" of a line equation. The 2 most common forms I see associated with a tangent line are:\[y-f(a)=f'(a)(x-a)\]or\[y=f'(a)(x-a)+f(a)\]
given that x=a=1 in this case, and f'(1)=27/4 ... yes, you will need to determine y=f(1) to finish it up

Answer 16

it might be useful to note that:\[y=f'(a)(x-a)+f(a)\\y=f'(a)x-f'(a)a+f(a)\]
\[y=f'(1)x-f'(1)+f(a)\]would be the usual y=mx+b from algebra

Answer 17

f(1), not f(a) ... lol, thats what i get for trying to look smart

Answer 18

it's okay your smart i'm the one that is not so smart today. but...

Answer 19

i though i would plug in the slope to the original equation to determine a value for y and then plug in x, y and the slop to get be and then write my new equation. but i'm not getting an answer on my sheet that's an option

Answer 20

f(x) = y = x^3*sqrt(x^3+3); x=1

f(1) = y = 1^3*sqrt(1^3+3)
= sqrt(4)
= 2

so the given point is (1,2) for the line equation to work out ;)

Answer 21

yes and then i got -19/4 as b

Answer 22

i made an algebra mistake again. that's why i usually do everything twice.

Answer 23

\[y-2=\frac{27}{4}(x-1)\]
\[y=\frac{27}{4}x-\frac{27}{4}+2\]
\[y=\frac{27}{4}x-\frac{27+8}{4}\]

Answer 24

the first one is a usual tangent line format .. point slope form

the last one is it swapped about into slope intercept form

Answer 25

oh okay

Answer 26

thank you so much

Answer 27

youre welcome, ive got real world job stuff to take care of for a little bit ... so i wont be around :) but there are plenty of people here that are smarter than me to help you out

good luck

Answer 28

well i really appriciate it and have a good day