Question

What does it mean if it wants me to find du/dx? of a function?

Answer 1

Find the derivative of u with respect to x .

Answer 2

So what if it's saying it wants me to find du/dx of the function \[y=(x+1)^2(x+3)^4(x+5)^6\] and it wants me to make \[u=\ln y\]

Answer 3

it wants you take ln of both sides of the original equation

Answer 4

first take ln of both sides, we are going to take derivative of simplified expression
$$ \Large {
y=(x+1)^2(x+3)^4(x+5)^6
\\ \iff \\
\ln(y)=\ln((x+1)^2(x+3)^4(x+5)^6)

} $$

Answer 5

the right side can be simplified

Answer 6

first take ln of both sides, we are going to take derivative of simplified expression
$$ \large {
y=(x+1)^2(x+3)^4(x+5)^6
\\ \iff \\
\ln(y)=\ln((x+1)^2(x+3)^4(x+5)^6)
\\ \iff \\

\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\ \iff \\
\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\\ln(y)= 2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)

} $$

Answer 7

the latter equation is much easier to take derivative of

Answer 8

So would I use the product rule to take the derivative?

Answer 9

now it is not necessary to use product rule, before it was

Answer 10

first take ln of both sides, we are going to take derivative of simplified expression
$$ \large {
y=(x+1)^2(x+3)^4(x+5)^6
\\ \iff \\
\ln(y)=\ln((x+1)^2(x+3)^4(x+5)^6)
\\ \iff \\

\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\ \iff \\
\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\\ln(y)= 2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)
\\~~\\

\frac{d}{dx}(\ln(y))= \frac{d}{dx}[2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)]
} $$

Answer 11

I got \[\frac{ 2 }{ (x+1) }+\frac{ 4 }{ (x+3) }+\frac{ 6 }{ (x+5) }\]

Answer 12

correct, thats for the right side, and the left side is ?

Answer 13

I'm not sure what the derivative of ln(y) is

Answer 14

Is it the same as the derivative of y=ln(x)?

Answer 15

almost, we use a chain rule here

Answer 16

I am getting 1/y x 1= 1/y

Answer 17

$$ \Large {
\frac{d}{dx}f(y(x)) = \frac{df}{dy}\cdot \frac{dy}{dx}

} $$

Answer 18

$$ \Large {
\frac{d}{dx}f(y(x)) = \frac{df}{dy}\cdot \frac{dy}{dx}
\\ \therefore \\
\frac{d}{dx}\ln(y(x)) = \frac{1}{y}\cdot \frac{dy}{dx}


} $$

Answer 19

Okay, thanks

Answer 20

$$
\large {
y=(x+1)^2(x+3)^4(x+5)^6
\\ \iff \\
\ln(y)=\ln((x+1)^2(x+3)^4(x+5)^6)
\\ \iff \\

\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\ \iff \\
\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\\ln(y)= 2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)
\\~~\\
\frac{d}{dx}(\ln(y))= \frac{d}{dx}[2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)]
\\\frac{1}{y}\cdot \frac{dy}{dx}=\frac{ 2 }{ (x+1) }+\frac{ 4 }{ (x+3) }+\frac{ 6 }{ (x+5) }
\\ \frac{dy}{dx}=y \cdot \left( \frac{ 2 }{ x+1 }+\frac{ 4 }{ x+3 }+\frac{ 6 }{ x+5 } \right)
}
$$

Answer 21

$$
\large {
y=(x+1)^2(x+3)^4(x+5)^6
\\ \iff \\
\ln(y)=\ln((x+1)^2(x+3)^4(x+5)^6)
\\ \iff \\

\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\ \iff \\
\\\ln(y)= \ln((x+1)^2) + \ln((x+3)^4) + \ln((x+5)^6)
\\\ln(y)= 2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)
\\~~\\
\frac{d}{dx}(\ln(y))= \frac{d}{dx}[2\ln(x+1) + 4\ln(x+3) + 6\ln(x+5)]
\\\frac{1}{y}\cdot \frac{dy}{dx}=\frac{ 2 }{ (x+1) }+\frac{ 4 }{ (x+3) }+\frac{ 6 }{ (x+5) }
\\ \frac{dy}{dx}=y \cdot \left( \frac{ 2 }{ x+1 }+\frac{ 4 }{ x+3 }+\frac{ 6 }{ x+5 } \right)
\\
\frac{dy}{dx} =\ln((x+1)^2(x+3)^4(x+5)^6)\cdot \left( \frac{ 2 }{ x+1 }+\frac{ 4 }{ x+3 }+\frac{ 6 }{ x+5 } \right)

}
$$