Question

how to get dy/dx for (2^-y) + (2^-x)=1

Answer 1

HINT: Logarithm

Answer 2

is it can ther was a "+' ???

Answer 3

is the answer -1??

Answer 4

Wait, we want to derivative y, so first we have to isolate y. Can you do it?

Answer 5

It's going to be ugly, trust me.

Answer 6

mmm i hv a question ..
i think only include ln when "*" or "/" is it incorrect??

Answer 7

Yeah.
ln(a·b) = ln a + ln b
ln(a/b) = ln a - ln b

I forgot about that, that's why I told you to do something else.

Answer 8

is it incorrect?

Answer 9

No, you are correct.

Answer 10

so how inlude ln here?

Answer 11

Forgot ln, it's probably won't help.

To solve your problem, either derivative both sides then isolate y' or isolate y then derivative it. Can you do either of them?

Answer 12

can u tell me a way ?

Answer 13

Either derivative both sides then isolate y' or isolate y then derivative it.

Can you do one of them?

Answer 14

so -yln(2)-xln(2) =0
y=-x
dy/dx= -1
is it correct?

Answer 15

incorrect??

Answer 16

No, there is no numerical answer. We have two unknown variables.

We just need to find y' or dy/dx

Answer 17

2^-y +2^-x =1
ln{2^-y +2^-x }=ln 1
-2^y ln(2)-2^xln(20=0
2^y=-2^x
y=-x
dy/dx=-1
???????

Answer 18

derivative of a^x is ?

Answer 19

u need to know that formula and CHAIN rule,....
u know chain rule ??

Answer 20

wow

Answer 21

what @hartnn said
you need both \(\frac{d}{dx}b^x=b^x\ln(b)\) and the chain rule to get \(\frac{d}{dx}2^{-x}=-2^{-x}\ln(2)\)

Answer 22

can u explain it?

Answer 23

yes

Answer 24

the derivative of \(b^x\) is \(b^x\ln(b)\) so for example the derivative of \(10^x\) is \(10^x\ln(10)\)
i can explain why this is true if you like

Answer 25

is it {2^(x+y) -2^x} / {2^y - 2^ (x+y) }

Answer 26

is it the answer?

Answer 27

to know why this is true, you need to know that \(b^x=e^{x\ln(b)}\) and then you can take the derivative using the chain rule, you get \(e^{x\ln(b)}\times \ln(b)\) which is the same as \(b^x\ln(b)\)

Answer 28

i doubt it, but we can check
start with
\[ 2^{-y} + 2^{-x}=1\] take derivative wrt \(x\) and get
\[-\ln(2)2^{-y}y'-\ln(2)2^{-x}=0\]

Answer 29

now solve for \(y'\)
you do not take the log of the whole thing which i think is confusing you
you cannot "simplify" \(2^{-y}+2^{-x}\) by taking the log . it is a sum, not a product

Answer 30

how u include ln ? there was "+"????

Answer 31

i think there are a few things confusing here
lets start with the basics
do you know what the derivative of \(2^{-x}\) is? you need that one

Answer 32

2^x is 2^x ln(2)
2^y is 2^y dy/dx ln(20
2^(x+y) is 2^(x+y) (1+dy/dx) ln(2)

Answer 33

ok but you do not have \(2^{x+y}\) in the problem

Answer 34

so 2^x + 2^y dy/dx = 2^(x+y) (1+dy/dx)

Answer 35

the original problem is this;
2^x +2^y = 2^(x+y)

Answer 36

i thought you had \[2^{-y}+2^{-x}=1\] ??

Answer 37

its correct too
it was divide by 2^(x+y)

Answer 38

ok then lets start with what you wrote above
do you know the derivative of \(2^{-x}\)?

Answer 39

it is -x* 2^-x * ln(2)

Answer 40

no it is \(-\ln(2)2^{-x}\)

Answer 41

oh yes not x sorry

Answer 42

so
-ln2^-2 -dy/dx ln(2)2^-y =0

Answer 43

so then we are ready to go
you get
\[-\ln(2)2^{-y}\times y'-\ln(2)2^{-x}=0\]solve for \(y'\)

Answer 44

dy/dx= 2^-x / 2^-y ?????

Answer 45

off by a minus sign

Answer 46

oh yes

Answer 47

but the given answer is;
{2^(x+y) - 2^x} / {2^y - 2^(x+y)}

Answer 48

how about this: \[2^{-y}+2^{-x}=1\] or \[2^x+2^y=2^{x+y}\] after removing negative exponents.
now apply the derivative to both sides.