Question

let f be a function such that f'(x) = (x^5+1)^-1 and y=f(x^2). Find dy/dx and the tangent to y=f(x^2) at x=1

Answer 1

what does f(x^2) mean...

Answer 2

Sam, a girl needs some help here. gimme somma DAT IQ

Answer 3

done.

Answer 4

Or
\[y=f(x^2) \\ \\ \frac{dy}{dx}=f'(x^2) \\ \\ \frac{dy}{dx}((x^2)^5+1)^{-1}\]

Answer 5

\[\int\limits \frac{1}{(x^5+1)}dx= \ln(x^5+1)+c\]

Answer 6

thanks that helps a lot

Answer 7

\[\frac{dy}{dx}\color{red}{=}(x^{10}+1)^{-1}\]

Answer 8

wow what math is this...calc1?

Answer 9

whoops sorry wrong info jhanny! Not integrate :P

Answer 10

Engineering maths 1

Answer 11

oh... then \[\frac{ dy }{ dx }=(x^{10}+1)^{-1}\]\[dy = (x^{10}+1)^{-1}dx\]\[\int\limits dy=\int\limits (x^{10}+1)^{-1}dx\]\[y= \ln(x^{10}+1) +c\]

Answer 12

Is that.....proper?

Answer 13

Something like a... separable equation or implicit differentiation... i forgot the name of that. Maybe...idk if i'm wrong.

Answer 14

we have \[\frac{dy}{dx}=(x^{10}+1)^{-1}\]
already, just substitute x=1 to find tangent point

Answer 15

Oh nvm -_-

Answer 16

sorted. Thanks a lot. was stumped on that one

Answer 17

oh,so you have the slope,given by that equation,and you're trying to find instantaneous velocity at that point? I mean that's what a tangent point would mean, right?