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 October 6th, 2012, 08:03 PM #1 Newbie   Joined: Oct 2012 Posts: 15 Thanks: 0 Calc How to find derivative of y when y=4sin(sq rt x)-2(sq rt x)cos(sq rt x)
 October 6th, 2012, 08:32 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Calc We are given: $y=4\sin$$\sqrt{x}$$-2\sqrt{x}\cos$$\sqrt{x}$$$ To differentiate, we will need to use the chain and product rules: $\frac{dy}{dx}=4\cos$$\sqrt{x}$$$$\frac{1}{2\sqrt{x }}$$-$$2\sqrt{x}\(-\sin\(\sqrt{x}$$$$\frac{1}{2\sqrt{x}}$$\)+$$\frac{ 1}{\sqrt{x}}$$\cos$$\sqrt{x}$$\)=$ $\frac{2}{\sqrt{x}}\cos$$\sqrt{x}$$+\sin$$\sqrt{x}\ )-\frac{1}{\sqrt{x}}\cos\(\sqrt{x}$$=\frac{1}{\sqrt{ x}}\cos$$\sqrt{x}$$+\sin$$\sqrt{x}$$$
 October 6th, 2012, 08:42 PM #3 Newbie   Joined: Oct 2012 Posts: 15 Thanks: 0 Re: Calc find 2nd derivative of, f'', of f when f(x)= (x+2)/(x-3)
 October 6th, 2012, 08:47 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Calc Use the quotient rule, twice. Post your attempt, and if you get stuck, I will help.
October 6th, 2012, 11:59 PM   #5
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Joined: Aug 2012
From: New Delhi, India

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Re: Calc

Quote:
 Originally Posted by henoshaile find 2nd derivative of, f'', of f when f(x)= (x+2)/(x-3)
$f(x)=\frac{x-3+5}{x-3}
f(x)=1+\frac{5}{x-3}$

Now it is easier to apply your standard chain rule. I have noticed that sometimes quotient rule becomes too complicated.
Therefore, it is much better to prefer algebraic reduction into easier functions.
Regards,
Rejjy
07-Oct-2012
12:28 IST

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