Question

Help with derivatives of natural logs please?

Answer 1

first of all, this is not simply derivatives of natural logs. it's logarithmic differentiation which is a bit more involved.

the idea is to rewrite by taking logs and then using the properties of logs to simplify the differentiation.

in the first one, you only need to rewrite the expression using the properties of logs...

\[\ln \frac{ e^{2x}\sqrt[3]{x-4x^5} }{ (x^3-6x)^7 }=\ln{e^{2x}}+\ln{\sqrt[3]{x-4x^5}}-\ln{ (x^3-6x)^7}\]
\[\Rightarrow y=2x+\frac{ 1 }{ 3 }\ln{(x-4x^5)}-7\ln{(x^3-6x)}\]

now just differentiate

Answer 2

for the second, you can take the log of both sides and simplify...\[e^{x+y}=xy \Rightarrow x+y = \ln{xy} = \ln x + \ln y \Rightarrow y -\ln y= \ln x - x\]
upon differentiation, you get
\[y' - \frac{ y' }{ y }=\frac{ 1 }{ x }-1\Rightarrow y'\left( \frac{ y-1 }{ y } \right)=\frac{ 1-x }{ x }\Rightarrow y'=\frac{ 1-x }{ x }\cdot \frac{ y }{ y-1 } \]

Answer 3

Give the third one a shot...

Answer 4

Do I take the logs of both sides like the problem before?

Answer 5

yeah

Answer 6

So...
\[lnf(x)^y={lny}^{f(x)}\]
\[ lny*lnf(x)=lnf(x)*lny\]

Answer 7

hey see in it..
Question is e^x+y=xy

Answer 8

\[y \ln f(x) = f(x) \ln y\]

Answer 9

\[\frac{ y }{ f(x) }=\frac{ f(x) }{ y }\]

Answer 10

no, you'd want the y's together and the f(x)'s on the other side if possible

Answer 11

\[f(x)^y=y^{f(x)}\Rightarrow y \ln {f(x)} = f(x) \ln y \Rightarrow \frac{ y }{ \ln y }= \frac{ f(x) }{ \ln f(x) }\]

Answer 12

Apply quotient rule?

Answer 13

yes or you can take logs again

Answer 14

\[y-lny=f(x)-lnf(x)\]
Take the derivative of this?

Answer 15

\[\ln y - \ln \ln y = \ln f(x) - \ln \ln f(x)\]
you'd have to take the derivative of this