Help Differentiate Please? (1 Viewer)

Zokunu · Dec 12, 2013

a) 2x^2 + 1 / √x^2 +1

Step by Step please? I keep getting 2x^3 + 3x/ x^2 + 1......==''

Answer: 2x^3 - 3x / (x^2 + 1)^3/2

HeroicPandas · Dec 12, 2013

Step:
-Normal application of quotient rule
-Common denominate the numerator
-Simplify and put everything as 1 fraction

$$Let $y = \frac{2x^2 + 1}{\sqrt{x^2+1}} \\ \\ \frac{dy}{dx} = \frac{(4x)\sqrt{x^2+1}- \left ( \frac{x}{\sqrt{x^2+1}} \right )(2x^2+1)}{x^2+1} \\ \\ \frac{dy}{dx} = \frac{\left [ \frac{4x(x^2+1)-x(2x^2+1)}{\sqrt{x^2+1}} \right ]}{x^2+1} \\ \\ \frac{dy}{dx} = \frac{2x^3 \fbox{+} 3x}{(x^2+1)^{\frac{3}{2}}}$

and yeh...i also have the plus instead of minus

anomalousdecay · Dec 13, 2013

HeroicPandas said:
Step:
-Normal application of quotient rule
-Common denominate the numerator
-Simplify and put everything as 1 fraction

$$Let $y = \frac{2x^2 + 1}{\sqrt{x^2+1}} \\ \\ \frac{dy}{dx} = \frac{(4x)\sqrt{x^2+1}- \left ( \frac{x}{\sqrt{x^2+1}} \right )(2x^2+1)}{x^2+1} \\ \\ \frac{dy}{dx} = \frac{\left [ \frac{4x(x^2+1)-x(2x^2+1)}{\sqrt{x^2+1}} \right ]}{x^2+1} \\ \\ \frac{dy}{dx} = \frac{2x^3 \fbox{+} 3x}{(x^2+1)^{\frac{3}{2}}}$

and yeh...i also have the plus instead of minus

Seems right to me.

ChillTime · Dec 13, 2013

When in doubt, ask Wolfram.

http://www.wolframalpha.com/

anomalousdecay · Dec 13, 2013

ChillTime said:
When in doubt, ask Wolfram.

http://www.wolframalpha.com/

No. When in doubt, ask our one and only Heroic!

Help Differentiate Please? (1 Viewer)

Zokunu

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HeroicPandas

Heroic!

anomalousdecay

Premium Member

ChillTime

Member

anomalousdecay

Premium Member

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