Higher Derivatives problem?

Question

Need to find the 4th derivative of:

f(x) = sin(lnx)dx

I got to the second derivative but I do not think it is correct. This is what I got as the second derivative:

f''(x) = -sin(lnx)(x^-2)-cos(lnx)(x^-2)

Also, if anyone knows of a TI-83 program or online program that does derivative please let me know. I'm in calculus 2 and doing this sort of thing again has been a drag.

Answer 1

f(x) = sin(lnx)

f '(X) = sinx/xlnx

f ''(x)= (xlnxcosx -(lnx+1) sinx)/(xlnx)^2

go on in the same way.

Answer 2

You simply use the chain rule, which states that df(g(x))/dx = f'(g(x))*g'(x)

f(x) = sin(lnx)

f'(x) = cos(lnx)*(1/x) = cos(lnx)*x^-1

And now you use the multiplication rule, which states that

df(x)g(x)/dx = f'(x)g(x) + g'(x)f(x)

f'(x) = cos(lnx)*x^-1

f''(x) = -sin(lnx)*x^-1*x^-1 + cos(lnx)*-x^-2

       = -sin(lnx)*x^-2 - cos(lnx)*x^-2

In other words, you were doing it right.

f'''(x) = -cos(lnx)*x^-3 + 2sin(lnx)*x^-3 + sin(lnx)*x^-3 + 2cos(lnx)*x^-3

       = cos(lnx)*x^-3 + 3sin(lnx)*x^-3

f'''(x) = -sin(lnx)*x^-4 - 3cos(lnx)*x^-4 + 3cos(lnx)*x^-4 - 9sin(lnx)*x^-4

       = -10sin(lnx)*x^-4

^^^^^^^^^

I'm reasonably sure that's right, but I'd check it before I used it if I were you.

Answer 3

f'(x) = [cos(lnx)]/x

f''(x) = -[sin(lnx) + cos(lnx)]/x^2

f'''(x) =[ -xcos(lnx) - xsin(lnx) + 2x[sin(lnx) + cos(lnx)]]x^4