Question

find the slope of the tangent line: y=7x^1/2+x^3/2, x=25

Answer 1

first find the derivative of the line

Answer 2

(7x^5/2+24)/(2x^4)

Answer 3

what the...?

Answer 4

my bad trying to do two problems at once...

Answer 5

but the derivative is 82/10?

Answer 6

i doubt it

Answer 7

i didnt think so either but i cant find it

Answer 8

\[y=7x^{1/2}+x^{3/2}\]

is this your problem?

Answer 9

correct and x=25

Answer 10

dont care about the x=25

the derivative of x^n is n*x^(n-1)

Answer 11

\[y'=\frac{7}{2}x^{-1/2}+\frac{3}{2}x^{1/2}\]

plug in for x and solve for y'
and thats the slope at x=25

Answer 12

5.44341649?

Answer 13

\[y= 7x^\left( \frac{ 1 }{ 2 } \right)+ x^\left( \frac{ 3 }{ 2 } \right) , x=25\]
you mean this?

Answer 14

yes

Answer 15

\[x^{1/2} =\sqrt{x}\]
\[x^{-1/2}=\frac{1}{\sqrt{x}}\]

Answer 16

\[\frac{ dy }{ dx }= \frac{ 7 }{2 }x^\left( \frac{ -1 }{ 2 } \right) + \frac{ 3 }{ 2 }x^\left( \frac{ 1 }{ 2 } \right) ; x=25\]

Answer 17

i need the slope of the tangent line

Answer 18

dy/dx is the slope...

Answer 19

ya the slope would be \[\frac{ dy }{ dx }\]

Answer 20

I keep plugging x=25 into f'(x) and getting my original m=82/10.

Answer 21

it might be \[\frac{ 82 }{ 10 }\]

Answer 22

i tried that and it says incorrect

Answer 23

then either the question is wrong or you put the wrong question here ;)

Answer 24

ya idk why its not working