Recent content by fr33pl4gu3

Thanks, but i solve the problem, the f(x) will be the nominator of the log and the g(x) will be the denominator of the log.

yes, and f is ln (12x-5/9x-2), where g is (12x-5/9x-2), correct??

Q. f(x)=ln (12x-5/9x-2) So by using the chain rule, i can get: (-4/3)((9x-2)2/(12x-5)2) and by using the quotient rule, i can get the final answer, which is: (2(-36x-8)(-36x-15)2-2(-36x-15)(-36x-8)2) ------------------------------------------------------------------...

How to solve this derivative??y = x4 . 4x How to solve this derivative?? y = x4 . 4x

Is this correct: (12/12(ln9x-2)-5)-(9/9x-2)

Then, is this f(x) = ln (12x-5) g(x) = ln (9x-2) f'(x) = 12/12x-5 g'(x) = 9/9x-2 Correct??

f(x) = ln (12x-5/9x-2) f'(x) = (4/3) (1/ln10)(9x-2/12x-5) Is this correct??

can my last answer be simplify one more step?

so the answer would be 4cos4x(cos3x) -3 sin4x(sin3x) Correct??

Is it f'(x)=4cos(4x) ??

sin4x can be written in this way (4(sin x)) or not?

y = sin( 4 x ) cos( 3 x ) f(x) = sin4x g(x) = cos3x f'(x) = cos4x g'(x) = -sin3x And by using the product rule, i'll get: cos4x(cos3x) - sin4x(sin3x) Is the answer correct or can be simplify again??

Yes, got it. Thanks for pointing out the error.

Is it this way?? f'(x) = 15(5x-2)2 g'(x) = -8(2x+5)-5

y = ( 5 x-2 ) 3 / ( 2 x+5 ) 4 I do it this way: f(x) = (5x-2)3 g(x) = (2x+5)-4 f'(x) = 3(5x-2)2 g'(x) = -4(2x+5)-5 By using the product rule: [3(5x-2)2](2x+5)-4 + (5x-2)3[-4(2x+5)-5] 3(5x-2)2(2x+5)-4 -4(2x+5)-5(5x-2)3 What's wrong with this answer?