Differentiaton of Log help (1 Viewer)

[evilevoevil]

I need help differentiating these log q's

ln [x + sqrt(x^2+k)]

ln (x - 1/x)

ln [2x^3 sqrt (3-2x)]

ln [3x^2 sqrt (1-x) / (1 + x)^3]

Thanks in advance

 

[jules.09]

ln [x + sqrt(x^2+k)]

You just differentiate the inside and divide it by [x + sqrt(x^2+k)] i.e. the form of f'(x)/f(x)

ln (x - 1/x)

Apply log laws, where ln (a/b) = ln(a) - l(b)

ln [2x^3 sqrt (3-2x)]

It's also f'(x)/f(x), but this time you've got to use product rule as well.

ln [3x^2 sqrt (1-x) / (1 + x)^3]

Apply log laws, where ln (a/b) = ln(a) - l(b) and then differentiate i.e. the form of f'(x)/f(x)


Practise, my friend.

 

[lychnobity]

ln [3x^2 sqrt (1-x) / (1 + x)^3]

Thanks in advance

For questions like these, where you have to differentiate something like ln(fraction), you can use the log laws to simplify first before differentiating.

Eg.

ln [3x²√(1-x) / (1 + x)³] = ln [3x²√(1-x)] - ln(1+x)³

= ln(3x²) + 0.5ln(1+x) - 3ln(1+x)

=ln(3x²) - 2.5ln(1+x)

So the derivative is:

(2/x) -2.5(1/1+x)

 

[lychnobity]

ln [x + sqrt(x^2+k)]

ie 1/√(x² + k) EDIT: look at the table of standard integrals. You should see something of a similar form to the question.

ln (x - 1/x)

= 1 - (-x^-2)/ x - x^-1 = x² + 1/ x(x² -1)

ln [2x^3 sqrt (3-2x)]

d/dx ln 2x³ + 0.5ln (3-2x) = 3/x - 1/(3-2x)

 

[evilevoevil]

Thanks for the help

I also need help with this question:
ln (ln x)

 

[jet]

 

[oly1991]

u could also just do the f'(x)/f(x)

y=ln(lnx)
y'=1/x/(lnx)
= 1/xlnx

 

[evilevoevil]

Need some help integrating this question

(3+sqrt x)^2/20x dx

 

[study-freak]

Need some help integrating this question

(3+sqrt x)^2/20x dx

expand the numerator
(9+6sqrt x+x)/20x dx
=(9/20x+3/(10sqrt x)+1/20) dx

integrate it
9/20ln abs(x) +(3 sqrt x)/5+x/20+C