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I can't find out what I'm doing wrong again...

$$f(x)=\frac{x^2+4x+3}{\sqrt{x}}$$

$$f(x)=\frac{x^2}{\sqrt{x}}+\frac{4x}{\sqrt{x}}+\frac{3}{\sqrt{x}}$$

$$f(x)=x^2(x^{9-1/2}) + 4x(x^{-1/2})+3x^{-1/2}$$

$$f(x)=x^{3/2}+4x(x^{-1/2})+3x^{-1/2}$$

$$f'(x)=(x^{3/2}+4x(x^{-1/2})+3x^{-1/2})'$$

$$f'(x)=\frac3{2}x^{1/2}+(4x(x^{-1/2})'+x^{-1/2}(4x)')-\frac32x^{-3/2}$$

$$f'(x)=\frac{3}2x^{1/2}-2x^{-3/2}+4x^{-1/2}-\frac{3}{2}x^{-3/2}$$

$$f'(x)=\frac{3}2x^{1/2}+2x^{-1/2}-\frac72x^{-3/2}$$

$$f'(x)=\frac{3}2\sqrt{x}+\frac{2}{\sqrt{x}}-\frac{7}{2\sqrt{x^3}}$$

When the answer in the book is:

$$f'(x)=\frac{3}2\sqrt{x}+\frac{2}{\sqrt{x}}-\frac3{2x\sqrt{x}}$$

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3  
Using the chain rule on $x\cdot x^{-1/2}$ instead of simplifying to $\sqrt{x}$ first is not reccomendable! –  Winther Jul 11 at 1:54
1  
In line 7, a $x$ is dropped. –  Mohammad Khosravi Jul 11 at 1:57
    
Yeah, that's where I went wrong. I should have just combined the x and x^{-1/2}, plus the dropped x. Thanks for the help –  KKendall Jul 11 at 1:58

2 Answers 2

up vote 1 down vote accepted

$$\begin{align} f'(x)&=\frac{d}{dx}\frac{x^2+4x+3}{\sqrt{x}} \\ &= \frac{d}{dx}\left(\frac{x^2}{\sqrt{x}}+\frac{4x}{\sqrt{x}}+\frac{3}{\sqrt{x}}\right) \\ &=\frac{d}{dx}\left(x^{\frac{3}{2}}+4\sqrt{x}+3(x)^{-\frac{1}{2}}\right) \\ &=\frac{3}{2}x^{\frac{1}{2}}+\frac{4}{2\sqrt{x}}-\frac{3}{2x^{\frac{3}{2}}}. \end{align}$$ This is simplified as $$f'(x)=\frac{3}{2}\sqrt{x}+\frac{2}{\sqrt{x}}-\frac{3}{2x^{\frac{3}{2}}}.$$

Note. The term $x^{\frac{3}{2}}$ can be expressed as $x^1\cdot x^{1/2}=x\sqrt{x}.$

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Tip: convert the square roots to exponent form, and combine exponents before taking the derivative.

$$f(x)=\frac{x^2+4x+3}{\sqrt{x}}$$

$$f(x)=(x^2+4x+3)(x^{-1/2})$$

$$f(x)= x^{3/2}+4x^{1/2}+3x^{-1/2}$$

$$f'(x)=\frac 3 2 x^{1/2}+2x^{-1/2}-\frac 3 2 x^{-3/2}$$

$$f'(x)=\frac {3\sqrt{x}} 2 + \frac 2{\sqrt{x}}-\frac 3 {2 x\sqrt{x}}$$

$$f'(x)=\frac {6x^2+ 4x-3 }{2 x\sqrt{x}}$$

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