What is the derivative of y = log16 ex - log4 sixth square root(x)?

[abel muroi]

I was given this problem

y = log₁₆ e^x - log₄ sixth square root(x)

I know the derivative of log₄ sixth square root(x) , it's just (1)/(6x log 4).

But i'm a little confused about the derivative of log₁₆ e^x .. if i apply the law of logarithms, then i get x log₁₆ e... but i don't know what to do next..

Should i apply the product rule of derivatives? Can anyone help me?

 

[Harry_the_cat]

I was given this problem

y = log₁₆ e^x - log₄ sixth square root(x)

I know the derivative of log₄ sixth square root(x) , it's just (1)/(6x log 4).

But i'm a little confused about the derivative of log₁₆ e^x .. if i apply the law of logarithms, then i get x log₁₆ e... but i don't know what to do next..

Should i apply the product rule of derivatives? Can anyone help me?

Use the log change of base rule to rewrite $\displaystyle log_16 e^x$ and then try to differentiate.

 

[pka]

I was given this problem
$\displaystyle y = \log_{16} (e^x) -\log_4(\sqrt[6]{x})$

$\displaystyle y' = \dfrac{1}{\log(16) } -\dfrac{1}{6x\log(4)}=\dfrac{1}{4\log(2) } -\dfrac{1}{12x\log(2)}$

 

[abel muroi]

$\displaystyle y' = \dfrac{1}{\log(16) } -\dfrac{1}{6x\log(4)}=\dfrac{1}{4\log(2) } -\dfrac{1}{12x\log(2)}$

Hmm..

I thought the derivative of log ₁₆ e^x was just log₁₆ e..

All i did was apply the law of logarithms to get (x) log ₁₆ e and then i tried to find the derivative of both terms.

I guess log₁₆ e doesn't really have a derivative, so i only took the derivative of x and got log₁₆ e as an answer.


Is this the correct method of finding the derivative of the function?

 

[Harry_the_cat]

Hmm..

I thought the derivative of log ₁₆ e^x was just log₁₆ e..

All i did was apply the law of logarithms to get (x) log ₁₆ e and then i tried to find the derivative of both terms.

I guess log₁₆ e doesn't really have a derivative, so i only took the derivative of x and got log₁₆ e as an answer.


Is this the correct method of finding the derivative of the function?

Yes that's a correct way of doing it, since log₁₆ e is a constant. Note also that log₁₆ e = 1/log_e 16 as in pka's response.

 

[ksdhart2]

Hmm..

I thought the derivative of log ₁₆ e^x was just log₁₆ e..

All i did was apply the law of logarithms to get (x) log ₁₆ e and then i tried to find the derivative of both terms.

I guess log₁₆ e doesn't really have a derivative, so i only took the derivative of x and got log₁₆ e as an answer.

Is this the correct method of finding the derivative of the function?

log₁₆(e) does have a derivative. It's 0, because the derivative of a constant with respect to any variable is always 0. So, you're correct in applying the constant multiplier rule here. Pka's answer looks different than yours, but they're equal, if you use the change of base formula to change from log base 16 to the natural logarithm:

$\displaystyle log_{16}\left(e\right)= \frac{ln\left(e\right)}{ln\left(16\right)}=\frac{1}{ln\left(16\right)}$

 

[pka]

Hmm..
I thought the derivative of log ₁₆ e^x was just log₁₆ e..

You need to learn the basics before trying to do any of these!!
\(\displaystyle \begin{align*} \log_b(x) &= y\\x &= b^y\\\log(x) &=y\log(b)\\\dfrac{\log(x)}{\log(b)} &= y\\\color{blue}\log_b(x) &= \dfrac{\log(x)}{\log(b)}\end{align*}\)

That means $\displaystyle \log_{16}\left(e^x\right)=\dfrac{x\log(e)}{\log(16)}=\dfrac{x}{\log(16)}=\dfrac{x}{4\log(2)}$

Thus $\displaystyle \frac{d}{dx}\left(\log_{16}\left(e^x\right)\right)=\dfrac{1}{\log(16)}=\dfrac{1}{4\log(2)}$

 

[Steven G]

Hmm..

I thought the derivative of log ₁₆ e^x was just log₁₆ e..

All i did was apply the law of logarithms to get (x) log ₁₆ e and then i tried to find the derivative of both terms.

I guess log₁₆ e doesn't really have a derivative, so i only took the derivative of x and got log₁₆ e as an answer.


Is this the correct method of finding the derivative of the function?

Abel, log₁₆ e is just a constant. What is the derivative of 7x? Did you use the product rule? What is the derivative of x*27? Did you use the product rule? If you are using the product rule then please use it again and find the derivative of kx where k is a constant. Do you see a formula now???? And yes, log₁₆ e does have a derivative. What is it?