I've been stuck on this problem for a while now and just can't seem to figure it out. If anyone knows how to do the problem, the help would be much appreciated. I know the answer for the problem, I just can't figure out how to get there. Problem: Find f'(x), f''(x), and f'''(x). f(x)= (x^2 - 3x)/(4x+2) I found f'(x) using quotient rule: f'(x)=(4x^2 + 4x -6)/(4x+2)^2 I can't figure out how to solve for the second two derivatives though. The answers for the subsequent derivatives of f(x) are: f''(x)= 56/(4x+2)^3 f'''(x)= -672/(4x+2)^4 I assume that since the exponent of the denominator is increasing by one each time, that the extended power rule is being used somehow. Beyond that I don't know. Do I use quotient rule again for the next two derivatives or extended power rule or what? I can't seem to get the math to work. If anyone knows the steps to reach the answer, the help is much appreciated. Thanks in advance if you do. Only problem I haven't been able to solve so far, and it is bugging me.:bang:
Hi there, You can re-write f'(x) like so: f'(x) = (4x^2 + 4x - 6)(4x + 2)^(-2) Then using the product rule (derivative of f*g = f'*g + f*g'), you get f''(x) = (8x + 4)(4x + 2)^(-2) + (4x^2 + 4x - 6)(-2)(4x + 2)^(-3)(4) The second term (4x^2 + 4x - 6)(-2)(4x + 2)^(-3)(4) might look weird because we needed to take the derivative of (4x + 2)^(-2). So bring down the exponent (-2), multiply it by the original term with the exponent reduced by one, and multiply it by the derivative of the inside. f''(x) = ( (8x + 4) / (4x + 2)^2 ) + ( ( -8 * (4x^2 + 4x - 6) ) / (4x + 2)^3 ) f''(x) = ( (8x + 4)(4x + 2) - 8(4x^2 + 4x - 6) ) / (4x + 2)^3 Note we make the exponent positive for the denominator since it's being divided by again. f''(x) = ( 32x^2 + 16x + 16x + 8 - 32x^2 - 32x + 48 ) / (4x + 2)^3 f''(x) = 56 / (4x + 2)^3 f'''(x) = 56 * (4x + 2)^(-3) f'''(x) = 0 + (56)(-3)(4x + 2)^(-4)(4) f'''(x) = -672 / (4x + 2)^4 Hope that helps!
Thanks a bunch. I was missing the derivative of the denominator when bringing it up.Think I was bringing the exponent on the denominator to -3 before trying to solve for the derivative or something. Or rather, putting -3 out in front instead of -2. I had done multiple derivatives of a function before in the homework, but it had been something simple like f(x)=1/x^3. Thanks again, you saved me. Now time to do some derivatives of trigonometric functions. Seems pretty simple so far, but then so did this last assignment till this problem lol.
I think my brain just exploded a bit looking at that problem. You are awesome to be smart enough to even know what that means. I'm a social science person so that equation makes me feel queesy. Good luck on the rest of your hw. : )
ahhh calculus.... I'm sure I could have helped you if you asked this this past spring because I took calc last semester. Unfortunately my brain seemed to have erased all memory of calc and derivatives and shit
Totally agree. Social sciences, written language, all that sort of thing I'm fine with... anything above basic mathematics, particularly chicken scratches like the above, not so much
I'm glad others have already helped you with your homework because math was always my worst subject and looking at the step-by-step process sunandmoon did wants to give me a headache.
