Question

how do i find the equation of the tangent line of a given point for example x^1/4 at (1,1)

Answer 1

find the slope by finding y' and pluggin' 1 for x

Answer 2

\[f(x)=x^n => f'(x)=nx^{n-1}\]

Answer 3

to find the tangent line at \[(x_0,y_0)\]
slope=\[f'(x_0)=nx_0^{n-1}\]
equation of line is \[y-y_0=f'(x)(x-x_0)\]
since f'=nx^{n-1}
then we have
\[y-y_0=nx_0^{n-1}(x-x_0)\]

Answer 4

ok so for the equation of the tangent line for x^1/4 you would first find f'

Answer 5

yes as i did above

Answer 6

is that the only question?

Answer 7

yeah am just having problems solving it sorry

Answer 8

the example?

Answer 9

yeah

Answer 10

or you having trouble finding y'?

Answer 11

the example i think i got y' its1/4c3/4 isnt it

Answer 12

\[1/4x ^{3/4}\]

Answer 13

\[y'=\frac{1}{4}x^{\frac{1}{4}-1}=\frac{1}{4}x^{\frac{1}{4}-\frac{4}{4}}=\frac{1}{4}x^\frac{-3}{4}=\frac{1}{4x^{ \frac{3}{4}}} \]

Answer 14

\[f'(1)=\frac{1}{4(1)^\frac{3}{4}}=\frac{1}{4}\]

Answer 15

\[y-1=\frac{1}{4}(x-1)\]

Answer 16

see thats what i got but my book shows 1/4x+3/4

Answer 17

did you distribute and add 1 on both sides

Answer 18

\[y-1=\frac{x}{4}-\frac{1}{4}\]

Answer 19

\[y=\frac{x}{4}-\frac{1}{4}+1\]

Answer 20

\[y=\frac{x}{4}-\frac{1}{4}+\frac{4}{4}\]

Answer 21

\[y=\frac{x}{4}+\frac{3}{4}\]

Answer 22

yeah i see that now thanks a lot i appreciate it