help differentiating logarithms?

Question

I've sort of forgotten.
I know that y= logeX, dy/dx= 1/X
but with something like this y=loge(X^2+5) I'm a little stuck. Could someone please help me and show working out :)
And also, what are the differention formulas I need to know for logarithams?
Thanks!

Answer 1

it is 1/(x^2+5) multiplied by d/dx of (x^2+5) which is 2x
so the answer is 2x/(x^2+5)

Answer 2

In addition to the formula you already have

y = ln(x), dy/dx = 1/x

You need to generalize this using the chain rule:

y = ln(u(x)), dy/dx = [1/u(x)] du/dx

In your example, u(x) = x^2 + 5, so du/dx = 2x and using the above formula for y = ln(x^2 + 5):

dy/dx = [1/(x^2 + 5)](2x)

Answer 3

ok, we have y=ln(x^2+5) (ln mean the logarithm at base e)

we differentiate dy/dx = [1/(x^2+5)]*2x=2x/(x^2+5)

The key idea here is that we differentiate also the internal function (here it is x^2+5) and multiply by the differentiated function.

In general case if we have function f(u) and u is a function of x (u=u(x)), then we do the following df/dx = (df/du)*(du/dx).

Answer 4

Never forget the chain rule when you're working with Calculus.

if y = f(g(x)), then y' = f'(g(x))*g'(x).

Therefore, if y = ln(x^2 + 5), then y' = 1/(x^2 + 5)*2x = 2x/(x^2 + 5)

Answer 5

y = x^(lnx) take logs of the two facets lny = lnx^(lnx) use the log regulation logm^p = p×logm lny = lnx.lnx now differentiate a million/y dy/dx = lnx.(a million/x) + lnx (a million/x) a million/y dy/dx = (2lnx)/x dy/dx = y. (2lnx)/x dy/dx = (x^(lnx). 2lnx)/x dy/dx = (2lnx . x^(lnx))/x now utilizing index regulations dy/dx = (2lnx . x^(lnx).x ^(-a million) dy/dx = (2lnx . x^(lnx-a million) i'm hoping this helps

Answer 6

If G(x) = ln(f(x))

G'(x)

.. f'(x)
= ----
...f(x)