Question

y+xcosy=(x^2)y
f'X=?

Answer 1

@amistre64

Answer 2

implicit derivative i assume? or is this partial derivative?

Answer 3

I have no idea :/

Answer 4

i have to wonder about the notation used in the post. \(f_x\) generally denotes a partial derivative

Answer 5

its f(x）

Answer 6

if you simply mean y' then its implicit

Answer 7

for starters, forget everything you think you know about how x and y act .... thats useless ideas.

y+xcosy=(x^2)y

y is a power rule; how would you take the derivative of y?
\[\frac d{dx}(y)=?\]

Answer 8

f(x)=x^4tan^(−1)(3x)
this

Answer 9

question sorry

Answer 10

Implicit Differentiation

Answer 11

... so f'(x) of that inverse tangent function?

Answer 12

recall the product rule: \(D_x(fg)=f'g+fg'\) correct?

Answer 13

yes

Answer 14

if we let f=x, what is f' equal?

Answer 15

prolly used a bad choice of letters for the product rule, but that can be overlooked :)

Answer 16

what is the first step to solve this

Answer 17

the first step is in getting the correct equation; is it:

\[f(x)=x^{4tan^{-1}(3x)}\]
OR
\[f(x)=x^{4}~{tan^{-1}(3x)}\]

Answer 18

the second equation

Answer 19

then that is simply a product of the terms: x^4
and: tan^(-1)(3x)

you know the power rule, and the second term is gonna employ the chain rule as well

Answer 20

4x^3? for the power and I am stuck with chain rules.
I have veen practing it but not improving :/

Answer 21

if we want to follow a systematic setup then maybe do this:
\[D_x(uv)=u'v+uv'\]
\[u=x^4~:~u'=? \]
\[v=tan^{-1}3x~:~v'=?\]

Answer 22

u=4x^3

Answer 23

u' = 4x^3 good

\[D_x(uv)=u'v+uv'\]
\[D_x(uv)=4x^3~tan^{-1}(3x)+x^4v'\]

Answer 24

v' is the hard one

Answer 25

the chain rule is such that we simply peel away the functions to form a product\[D_x(f(g(h(x))))=f'(g(h(x)))~~g'(h(x))~~h'(x)~~x'\]

Answer 26

\[D_x[tan^{-1}(3x)]=(tan^{-1})'(3x)~~(3x')\]
\[D_x[tan^{-1}(3x)]=3(tan^{-1})'(3x)\]

ill have to review for myself what the tan-1' is :)

Answer 27

ohh haha this is why they call it chain rules! it looks like a chain

Answer 28

let:

y=arctan(x)

tan(y)=x

sec^2(y) y' = 1

y' = 1/(sec^2(y)) ; and we know y=arctan(x)

y' = 1/(sec^2(arctan(x))



yeah, 1/(1+x^2)