Find f'(x):

Question

Find f'(x):

anonymous · Answer

$$3x ^{2}+10x+37\div \sqrt{x}$$

anonymous · Answer

I got something like this: 
$$6x \sqrt{x}+10\sqrt{x}-1.5x ^{1.5}+5x ^{1/2}+18.5x^ {-1/2}$$

anonymous · Answer

sorry, the above all divided by x

anonymous · Answer

f'(x), i tried using the quotient rule

anonymous · Answer

Sorry, the question is $$(3x ^{2}+10x+37)\div \sqrt{x}$$ find f'(x)

anonymous · Answer

It's easier if you just write out the three terms of f explicitly:
f(x) = (3x^2+10x+37)/sqrt(x)
Hence
$$f(x) = 3x^{3/2} + 10x^{1/2} + 37x^{-1/2}$$
Now just differentiate, term by term:
$$f'(x) = {9 \over 2} x^{1/2} + 5x^{-1/2} - {37 \over 2}x^{-3/2}$$

anonymous · Answer

http://mathway.com/math_image.aspx?p=SMB02FSMB03(3xSMB02ESMB032SMB02eSMB03+10x+37)(x)SMB02ESMB03SMB02FSMB031SMB102SMB02fSMB03SMB02eSMB03SMB10SMB02fSMB03+x?p=178?p=42