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 November 27th, 2010, 07:27 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 derivatives help [color=#0040FF]1 ) If v is a differentiable function of x . then d/dx(1/v) = [/color] d/dx =1v^2 . [color=#0040FF]2 ) If x = sin y , then dy/dx = [/color] Cosy.y' [color=#0040FF]3 ) If y^2 = x then dy/dx = [/color] = 2y = 1 [color=#0040FF]4 ) The derivative of y = e^t sint with respect t is [/color] e^t(sint + cost ) d/dx|x| = zero [color=#0040FF]5 ) Find the dy/dx of y = sinu , u = x^2 [/color] Cosu.2x
 November 27th, 2010, 08:22 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: derivatives help 1.) $\frac{d}{dx}\left(\frac{1}{v}\right)=\frac{d}{dx}\ left(v^{\small{-1}}\right)=-v^{\small{-2}}\frac{dv}{dx}$ 2.) $x=\sin y$ Implicitly differentiate with respect to x: $1=\cos y\cdot\frac{dy}{dx}\:\therefore\:\frac{dy}{dx}=\se c y=\sec\left(\arcsin x\right)=\frac{1}{\sqrt{1-x^2}}$ 3.) $y^2=x$ Implicitly differentiate with respect to x: $2y\frac{dy}{dx}=1\:\therefore\:\frac{dy}{dx}=\frac {1}{2y}=\pm\frac{1}{2\sqrt{x}}$ 4.) You did correctly! $\frac{d}{dx}|x|=\frac{d}{dx}\left(x^2\right)^{\sma ll{\frac{1}{2}}}=\frac{1}{2}\left(x^2\right)^{\sma ll{-\frac{1}{2}}}\cdot(2x)=\frac{x}{\left(x^2\right)^{ \small{\frac{1}{2}}}}=\frac{x}{|x|}$ 5.) $\frac{dy}{du}\cdot\frac{du}{dx}=\frac{dy}{dx}$ $\frac{dy}{du}=\cos u=\cos\left(x^2\right)$ $\frac{du}{dx}=2x$ $\frac{dy}{dx}=2x\cos\left(x^2\right)$ You did it correctly, but perhaps you should substitute for u in your final answer.
November 27th, 2010, 08:26 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: derivatives help

Hello, r-soy!

The first three are incorrect.
(I'll let someone else comment on the others.)

Quote:
 $\text{1) If }v\text{ is a differentiable function of }x\text{. find }\frac{d}{dx}\left(\frac{1}{v}\right)$
$\frac{d}{dx}\left(v^{-1}\right) \;=\;-v^{-2}\,\frac{dv}{dx}$

Quote:
 $\text{2) If }x \,=\, \sin y\text{, find }\frac{dy}{dx}$
$1 \:=\:\cos y\,\frac{dy}{dx} \;\;\;\Rightarrow\;\;\;\frac{dy}{dx} \:=\:\frac{1}{\cos y} \;=\;\sec y$

Quote:
 $\text{3) If }y^2 \,= \,x\text{, find }\frac{dy}{dx}$

$2y\,\frac{dy}{dx} \:=\:1 \;\;\;\Rightarrow\;\;\;\frac{dy}{dx} \:=\:\frac{1}{2y}$

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