Question

Can someone explain why taking the derivative of 3(e^x + e^-x) = 3(e^x - e^-x)

Answer 1

f(x) =
\[3(e^x +e ^{-x}) \]

f'(x) = \[3(e^x - e ^{-x}) \]

Answer 2

I would expand and derive term by term.

Answer 3

where you have e^(-x), apply the chain rule for the exponent, but whatever you have in the chain for the exponent, you multiply it on the outside, not inside the exponent.

Answer 4

I mean differentiate term by term, not derive. (I get critised when I use word derive in this sense)

Answer 5

\(\large\color{blue}{f(x)=3e^{x}-3e^{-x}}\)

Answer 6

so what is the derivative of e^{-x}?

Answer 7

\(\large\color{blue}{f'(x)=\frac{d}{dx}3e^{x}-\frac{d}{dx}3e^{-x}}\)

\(\large\color{blue}{f'(x)=3\frac{d}{dx}e^{x}-3\frac{d}{dx}e^{-x}}\)

\(\large\color{blue}{f'(x)=3e^{x}-3\frac{d}{dx}e^{-x}}\)

Answer 8

e^x remains e^x.

Answer 9

\(\large\color{blue}{\frac{d}{dx}e^{-x}=e^{\color{red}{-x}}\times\frac{d}{dx}(\color{red}{-x}) }\)

Answer 10

Ok so that would be \[-1 e ^{-x}\]

Answer 11

yes,

Answer 12

yes, which is \-e^(-x)

Answer 13

Ok so thats why...

Answer 14

Thank you!!!

Answer 15

So we get,
\(\large\color{blue}{ f'(x)=3e^x-(-3x^{-x}) }\)
\(\large\color{blue}{ f'(x)=3e^x+3x^{-x} }\)

Answer 16

Thanks I get it now :))))

Answer 17

So when the initial function is,
\(\large\color{blue}{ f(x)=3e^x-3x^{-x} }\)
then
\(\large\color{red}{ f'(x)=3e^x+3x^{-x} }\)

and if, the initial function is,
\(\large\color{blue}{ f(x)=3e^x+3x^{-x} }\)
then
\(\large\color{red}{ f'(x)=3e^x-3x^{-x} }\)

Answer 18

right, just the chain rule, derivative of -x, you multiply times -1.

Answer 19

Yep, learning this stuff one step at a time haha.Thanks! )))

Answer 20

anytime:)
I was once new to this too:)
But it gets easy to differentiate any function, unlike integration.