Question

If y = 5/2(e^(x/5) + e^(-x/5)), prove that y'' = y/25.

Answer 1

I expanded the bracket:
y = 5/2(e^(x/5)) + 5/2(e^(-x/5))

Answer 2

Then I took the first derivative:

Answer 3

y ' = 5x/10(e^(x/5)) - 5x/10(e^(-x/5))

Answer 4

Now how should I go about with getting the second derivative? Should I factor out the 5x/10 and then use product rule?

Answer 5

\[y'=5/2[e ^{x/5}(1/5)+e^{-x/5}(-1/5)\]

Answer 6

\[y'=(5/2)(1/5)[e ^{x/5}-e ^{-x/5}]\]

Answer 7

Does it make a difference if we were to expand first with 5/2, and then take the first derivaitve?

Answer 8

not at all

Answer 9

Because that is what I did and then I took the dy/dx of each of the two terms.

Answer 10

but y did u multiply with x/5?

Answer 11

instead it is 1/5 the coefficient of x/5

Answer 12

No sorry I did the following:
y = 5/2(e^{x/5}) + 5/2(e^{-x/5})
y '= 5/2(1/5)*(e^{x/5}) + 5/2(-1/5)(e^{-x/5})

Answer 13

correct

Answer 14

So then:
y = 1/2(e^{x/5} + e^{-x/5})

Answer 15

y' u mean?

Answer 16

yes sorry, y'

Answer 17

y' = 1/2(e^{x/5} + e^{-x/5})

Answer 18

n there must b negative between to terms

Answer 19

coz 1/5 is the common factor

Answer 20

y' = 1/2(e^{x/5} - e^{-x/5}), right?

Answer 21

yes:)

Answer 22

So then for y'' I get..

Answer 23

\[y''=1/2[e ^{^{x/5}}(1/5)-e ^{-x/5}(-1/5)]\]

Answer 24

Yes I got that, then I expanded, and I got:
y'' = 1/10(e^{x/5} + e^{-x/5})

Answer 25

\[y''=1/10[e ^{x/5}+e ^{-x/5}]\]

Answer 26

again no need to expand

Answer 27

\[e ^{x/5}+ e ^{-x/5}=2y/5\]

Answer 28

y''=y/25

Answer 29

How did you get e^x/5 + e^-x/5 = 2y/5?

Answer 30

OHHH

Answer 31

I think I know

Answer 32

got it?

Answer 33

I got it thank you so much. So when you are taking derivative of a function that has a constant which is multipled by a bracket that can be differentiated, you just take the derivative of the terms inside the bracket and multiply by the constant in front of it?

Answer 34

right:)

Answer 35

Ok awesome. Thank you for your help.

Answer 36

my pleasure