1. ## derivative

what is the derivative of -3/x^2 + 3lnx - 6e^x

2. Derivative of

:: bx^a = bax^(a-1)

:: aln(x) = a/x

:: ae^(x) = ae^(x)
-------------------------

Use them Derivative of

f(x) + g(x) + k(x)
is
f'(x) + g'(x) + k'(x)

3. so its -6e^x + 3/x.

i have problems w the first one becasue its a fraction.

now if i leave the fraction its -3/ 2x. but should i leave the fraction or not?

or is the end result -3/ 2x - 6e^x + 3/x ?

4. Originally Posted by makaveli89
what is the derivative of -3/x^2 + 3lnx - 6e^x

It can be done as

Derivative of

-3/x^2 = -3x^(-2)

is given by

-3*-2 x^(-3)
= 6 x^(-3)

-----------------------

= 6 x^(-3) -6e^x + 3/x.

5. Originally Posted by makaveli89
what is the derivative of -3/x^2 + 3lnx - 6e^x
Your formatting is ambiguous. I think the other replies have assumed that you meant the following:

. . . . . $-\frac{3}{x^2}\, +\, 3\ln(x)\, -\, 6e^x$

Is the above what you meant? Or did you mean "-3/(x^2 + 3ln(x) - 6e^x)", which, in LaTeX, looks like the following?

. . . . . $\frac{-3}{x^2\, +\, 3\ln(x)\, -\, 6e^x}$

When you reply, please show what you have tried so far, so we can "see" which part is causing confusion. Thank you!