Question

Use log differentiation to find g'(x) if
g(x) = e^(5x)cubert(x^(3)+9)/(cos^(2)(x))

Answer 1

Original equation:
\[\frac{ e^{5x} \sqrt[3]{x^{3}+9} }{ (\cos^2(x) )}\]

Answer 2

My question is more about the process of using the log properties before doing the actual derivative. When I get to the part where I have to use ln(y)=ln(e^5x)+ln(cubert(x^3+9)-ln((cos^(2)(x))
my question is does the \[\ln(\cos^2(x))\] become \[2\ln(\cos(x))\]
because of the fact that\[\cos^2(x)=(\cos(x))^2\]
and this log of prop.
\[\ln(x^y)=yln(x)\]
?

Answer 3

@JerJason use Quotient Rule

Answer 4

I'm reposting the equation that I was referring to in my last post,\[\ln(y)=\ln(e^{5x})+\ln(\sqrt[3]{x^3+9})-\ln(\cos^{2}(x))\]

Answer 5

YES,
\(\ln [\cos^2x] = 2\ln \cos x\)

Answer 6

Okay. Thats what I thought. So in that specific rule ln(x^y)=yln(x)
the "Y" is whatever is outside of the x right? So for the following functions:
\[\ln(\sqrt(x))\]
\[\ln((x^2+1)^2)\]
the x would be whats in the "()" and the y would be the exponent?
Would the same rule apply for something such as:
\[f(x)=ln(x^{3x})\]?

Answer 7

\(\huge \ln ☺^♥ =♥ \ln ☺ \)

now \(☺ , and~ ♥\)
can be ANY functions

Answer 8

\(\ln(\sqrt(x)) = 1/2 \ln x\)
\(\ln((x^2+1)^2) = 2 \ln((x^2+1)\)
\(\ln(x^{3x}) = 3x \ln x\)

Answer 9

\[g \left( x \right)=\frac{ e ^{5x} \sqrt[3]{x^3+9}}{ \cos ^2x }=\frac{ e ^{5x \left( x^3+8 \right)^{\frac{ 1 }{ 3 }}} }{ \cos ^2x }\]
\[\ln {g(x)}=\ln e ^{5x}+\frac{ 1 }{ 3 }\ln \left( x^3+9 \right)-2\ln \cos x\]
\[=5x +\frac{ 1 }{ 3 }\ln \left( x^3+9 \right)-2 \ln \cos x\] (lne=1)
Diff. with respect to x
\[\frac{ g \prime \left( x \right) }{ g \left( x \right) }=5+\frac{ 1 }{ 3 }\times3 x^2-2\frac{ -\sin x }{ \cos x }\]

Answer 10

correction

Answer 11

@hartnn
Ok so
\[\ln(e^{5x}) \]
\[\ln(x^{5}) \]
wouldn't work because e^(5x) and x^5 is all part of one function(the "x" in ln(x))?

Answer 12

\[\frac{ g \prime \left( x \right) }{ g \left( x \right) }=5+\frac{ 1 }{ 3 }*\frac{ 3 x^2 }{ x^3+9 }-2\frac{ -\sin x }{ \cos x }\]

Answer 13

\(\ln e^{5x} = 5x \ln e = 5x \\ \ln x^5 = 5 \ln x\)

Answer 14

there is nothing that won't work....if there is an exponent, it comes as a co-efficient of log

Answer 15

Ah ok, I see now. Thanks.

Answer 16

welcome ^_^
hope you can solve your question entirely now :)

Answer 17

Yes I think I can now that I have that clarification.