Question

differentiate y = ln (3 + 2^x)

Answer 1

the derivative of y, is just dy/dx.

for the right hand side though, you nee d a chain rule.

Answer 2

derive the natural log of (3+2^x) as if it was just ln(x),
saying it will be
1 / (3 +2^x) and then multiply times the derivative of the inner part. (times the derivative of (3+2^x).

Answer 3

you need to find the derivative of 2^x.

Answer 4

\[\frac{dy}{dx}=\frac{1}{3+2^x}\times\color{red}{ (\frac{d}{dx}~[3+2^x]~)}\]

Answer 5

the red part, is I am mutliplying times the derivative of the inside. The chain rule.

Answer 6

Hello sorry and thank you! I've always got a problem with chain rule. I got exactly like your equation so give me a minute to work it out!

Answer 7

you just need to find the derivative of 2^x.... please post your work here though. I need to see what you do and don't know.

Answer 8

looks like you don't feel comfortable deriving 2^x, do you need help doing that, or not ?

Answer 9

Yes please..

Answer 10

\[y=2^x\]\[\ln(y)=\ln(2^x)\]\[\ln(y)~=x \ln(2)\]\[\frac{dy}{dx}~\times\frac{1}{y}=\ln(2)\]\[\frac{dy}{dx}=y~\ln(2)\]\[\frac{dy}{dx}=2^x~\ln(2)\]

Answer 11

at my last step I just plugged in 2^x back for y (since we know that y=2^x from the beginning).

So the derivative of 2^X is ln(2) 2^X

Answer 12

So my final answer is dy/dx = 2^x ln (2)/ 3+2^x. Is this correct?