Question

Can someone help me by explaining how to solve logarithmic differentiation?

Answer 1

Happy to try, but would MUCH prefer to work on an actual example. Otherwise your question is much too broad. By the way, one does not "solve" log differentiation; one "applies" it.

Answer 2

thank you! \[-3\ln (8x)\]

Answer 3

I'm assuming that you have the function f(x)=-3ln (8x).
One approach (not the only approach) would be to re-write this
function as f(x)=-3 [ln 8 + ln x].

ln 8 is a constant. What is its derivative?
x is a variable. What is the derivative of ln x?

What is the derivative of ln 8x?

Multiply this result by -3 and you'll be done. You must label your result:

dy/dx = -3 [ ? ]

Answer 4

That's what I don't understand, the derivative of ln8x & lnx

Answer 5

Another way would be to use the chain rule. 8x is a function in its own right.

Thus, the derivative with respect to x of f(x) = -3 ln 8x

would be\[f'(x)=-3\frac{ d }{ dx }\ln (8x)\]

Answer 6

Note that if u is a separate function, then the derivative of y=ln u is\[\frac{ dy }{ dx }=\frac{ 1 }{ u }\frac{ du }{ dx }\]

Answer 7

Try it.

Answer 8

So would the final answer be \[\frac{ dy }{ dx }=-3[\frac{ 1 }{ 8x }]\] ?

Answer 9

Logarithmic differentiation is usually applied to a problem involving an exponential. Was this your original function? \(\large\rm y=(8x)^{-3}\) and you took the log of each side? \(\large\rm ln(y)=-3ln(8x)\)

Or you're just simply trying to find the derivative of the log function?

Answer 10

I was given \[f(x)=-3\ln (8x)\]

Answer 11

Ah ok fair enough :)\[\large\rm \frac{d}{dx}-3\ln(8x)=-3\frac{1}{8x}\frac{d}{dx}(8x)\]Don't forget your all-important `chain rule`!
Multiply by the derivative of the inner function.

Answer 12

Oh my goodness thank you!