Question

need help finding the derivative

Answer 1

the problem is \[\frac{ \tan \frac{ 1 }{ x } }{ \frac{ 1 }{ x } }\]

Answer 2

\(\large\color{black}{ \displaystyle \frac{\tan\left(\frac{1}{x}\right)}{\frac{1}{x}} }\)

\(\large\color{black}{ \displaystyle x\tan\left(\frac{1}{x}\right) }\)

Answer 3

You could also use the product rule if it is easier.

\[xtan(1/x)\]

Answer 4

now, the product rule

Answer 5

the answer is \[\frac{ \sec^2\frac{ 1 }{ x }*\frac{ -1 }{ x^2 } }{ \frac{ -1 }{ x^2 } }\]

Answer 6

i dont get how it got that

Answer 7

If I tell you the derivative of tan(x)=sec^2(x), does that help?

Answer 8

\(\large\color{black}{ \displaystyle x\tan\left(\frac{1}{x}\right) }\)

the answer is:
\(\large\color{black}{ \displaystyle 1\times \tan\left(\frac{1}{x}\right)+x\times \sec^2\left(\frac{1}{x}\right)\times \left(-\frac{1}{x^2}\right) }\)

Answer 9

using the chain rule, and I know that d/dx tan(x)= sec^2(x), tnx

Answer 10

yeah i know the derivative of tan is sec^2 but where do the negative ones come from

Answer 11

I am using the chain rule, because it is tangent of (x^(-1) )

Answer 12

and why are they squared

Answer 13

d/dx x^(-1) = (-1)x^(-1-1) = -1/x^2

Answer 14

and sec^2(1/x) is the derivative of tan(1/x) without the chain rule

Answer 15

and with the chain rule,

d/dx tan(1/x) = sec^2(1/x) * (-1/x^2)

Answer 16

so, that component (derivative of tan(1/x) ) is used in the product rule

Answer 17

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left(x\times \tan\left(\frac{1}{x}\right)~~\right) ~~~= }\)

\(\large\color{black}{ \displaystyle 1\times \tan\left(\frac{1}{x}\right)+x\times \sec^2\left(\frac{1}{x}\right)\times \left(-\frac{1}{x^2}\right) }\)

Answer 18

is it suppose to be sec^2(1/x) * tan(-1/x^2) then or did something happen to the tan?

Answer 19

no, we aren't changing the angle

Answer 20

lets look at the derivative of tan(1/x) again, ok?

Answer 21

I supposed one could \(\bf \cfrac{ tan \left( \frac{ 1 }{ x } \right) }{ \frac{ 1 }{ x } }\implies tan \left( \frac{ 1 }{ x } \right)\cdot \left( \frac{ 1 }{ x } \right)^{-1}\implies tan \left( \frac{ 1 }{ x } \right)\cdot \left( \frac{ x }{ 1 } \right)^{+1}
\\ \quad \\
tan \left( \frac{ 1 }{ x } \right)\cdot x\implies tan(x^{-1})\cdot x\)

and apply the chain-rule to that, along witht he product rule

Answer 22

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left( \tan\left(\frac{1}{x}\right)~~\right) ~~~= }\)

\(\large\color{black}{ \displaystyle \color{blue}{\sec^2\left(\frac{1}{x}\right)}\times \color{red}{\frac{d}{dx}\left[\frac{1}{x}\right]} }\)

\(\large\color{black}{ \displaystyle \sec^2\left(\frac{1}{x}\right)\times \color{red}{\left(-\frac{1}{x^2}\right)} }\)

Answer 23

the blue is the derivative of the entire argument and in red we are taking the chain rule

Answer 24

okay

Answer 25

so,

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left( \tan\left(\frac{1}{x}\right)~~\right) ~~~=~~~\frac{-1}{x^2}\sec^2\left(\frac{1}{x}\right) }\)

Answer 26

and then the use of product rule

Answer 27

so your using the chain rule instead of the quotient rule

Answer 28

no

Answer 29

the chain rule is for the angle of 1/x
(because it is not just tan(x), but rather tan(1/x) )

the quotient rule we didn't use, not because we used the chain rule, but because we re-write what you had initially as x tan(1/x) and then used the product rule.

Answer 30

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left( x\tan\left(\frac{1}{x}\right)~~\right) ~=~\color{magenta}{1}\times \tan\left(\frac{1}{x}\right)+x\times\color{magenta}{\frac{-1}{x^2}\sec^2\left(\frac{1}{x}\right)} }\)

Answer 31

(in pink are the derivatives )

Answer 32

and all that is remaining to do, is to simplify it

Answer 33

wait where did the x in front of the tan(1/x) come from?

Answer 34

that is your initial function